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Really nice inequality. The answer is $C=\frac{1}{8}$ and the equality holds when two $x_i$ are equal and the rest are zero. We have $(\sum x_i)^4=(\sum x_i^2+2\sum x_ix_j)^2\geq 4(\sum x_i^2)(2\sum x_ix_j)= 8 \sum (x_ix_j\sum x_k^2)\geq 8\sum x_ix_j(x_i^2+x_j^2).$ The second ineq has equality if and only if $n-2$ of $x_i$ are $0$. So WLOG assume that $x_3=x_4=...=x_n=0$. Then for the first ineq to be equality we require that $x_1=x_2$ However that isclearly also sufficient for equality and we are done

Please stop copying entire solutions, Silouan, like you did with the problem about the hexagon, and also in this case ("your" solution comes straight from http://www.kalva.demon.co.uk/imo/isoln/isoln992.html, although you changed a few words).

Firstly I didnot say that this is ''my'' solution. And this is posted many days ago. And I consired good to post a solution. With this solution i)someone could insrired to find another solution ii) this problem doesnot seen as unsolved in the regional contests iii) I remind this beautiful problem to all. But if you want not me post such solutions then ok. It will not repeated. thanks for the hearing. Silouanos

There exist another solution for this. We WLOG assume that $x_1+.....+x_n=1$ and we have $\sum_{i=1}^n {x_i^3(1-x_i)}\leq C$ Now let the function $f(x)=x^3(1-x)$ . From Karamata inequality we have that $f(x_1)+....+f(x_n)\leq f(x_1)+f(1-x_1)+(n-2)f(0)\leq\frac{1}{8}$

silouan wrote: There exist another solution for this. We WLOG assume that $x_1+.....+x_n=1$ and we have $\sum_{i=1}^n {x_i^3(1-x_i)}\leq C$ Now let the function $f(x)=x^3(1-x)$ . From Karamata inequality we have that $f(x_1)+....+f(x_n)\leq f(x_1)+f(1-x_1)+(n-2)f(0)\leq\frac{1}{8}$ $f(x)=x^3(1-x)$ $f'(x)=3x^2 - 4x^3$ $f''(x)=6x-12x^2 \geq 0$ for $x \in \left [0, \frac{1}{2} \right ]$ it just concave for $x> \frac{1}{2}$

It suffices to prove it at the interval $[0,\frac{1}{2}]$.

the array $(x_1;x_2;x_3;...;x_n) \succ (x_1;1-x_1;0;...;0)$ , $f(x)$ is convex in $x \in \left [ 0, \frac{1}{2} \right ]$ then we have $f(x_1)+f(x_2)+...+f(x_n) \geq f(x_1)+f(1-x_1)+(n-2)f(0)$, did you reserve the inequality sign ?

orl wrote: Let $ n \geq 2$ be a fixed integer. Find the least constant $ C$ such the inequality \[ \sum_{i < j} x_{i}x_{j} \left(x^{2}_{i} + x^{2}_{j} \right) \leq C \left(\sum_{i}x_{i} \right)^4 \] holds for any $ x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $ \binom{n}{2}$ summands). For this constant $ C$, characterize the instances of equality. I will prove that $ C = \frac{1}{8}$. The inequality reduces to: $ 8\left (\sum_{i=0}^{n}x_{i}\right )\left (\sum_{i=0}^{n}x_{i}^{3}\right )\le\left (\sum_{i=0}^{n}x_{i}\right )^{4}+8\sum_{i=0}^{n}x_{i}^{4}$ It's obvious when $ x_1=..=x_n=0$. Assume that not all are zero. Let $ y_i = \frac{x_i}{x_1+x_2+...+x_n}, i=1,2,..,n$. Then we have to prove that: $ \sum_{i=1}^n 8y_i^3 - 8y_i^4 \le 1 \iff$ Consider $ f(x) = 8x^3 - 8x^4$. Then: $ f'(x) = 24x^2 - 32x^3$ $ f''(x) = 48x - 92x^2 = 48x(1-2x)$. So $ f''(x) \ge 0 \forall x \in [0;\frac{1}{2}]$ Wlog assume $ y_1 \ge y_2 \ge y_3 \ge ... \ge y_n$. Then: $ y_2,y_3,..,y_n \in [0;\frac{1}{2}]$. Since $ (y_2+y_3+..+y_n,0,0,...,0) \succ (y_2,y_3,...,y_n)$. We see that $ f(y_2+y_3+...+y_n) + f(0) + f(0) +.. + f(0) \ge f(y_2) + f(y_3) + .. + f(y_n)$. Let $ Z = y_2+..+y_n, Y = y_1$. Then since $ f(0) = 0$ we just have to prove that: $ f(Y) + f(Z) \le 1$ when $ Y+Z = 1$. But this is equivalent to: $ 8(Y+Z)(Y^3+Z^3) \le (Y+Z)^4 + 8Y^4 + 8Z^4 \iff$ $ (Y-Z)^4 \ge 0$. So equality iff $ x_1=x_2, x_3=x_4=...=x_n=0$ or permutations. QED

Wait... if they say "fixed" $n$, doesn't that mean that $C$ can be in terms of $n$? If so, I got that $C=\frac{n-1}{n^{3}}$ (where the largest value that this can attain is precisely $\frac{1}{8}$

huang wrote: Wait... if they say "fixed" $n$, doesn't that mean that $C$ can be in terms of $n$? If so, I got that $C=\frac{n-1}{n^{3}}$ (where the largest value that this can attain is precisely $\frac{1}{8}$ Yes. $C$ should be in terms of $n$. If $x_1=x_2=1,x_3=x_4 =\cdots=x_n$ we get $2^4C \ge 2 \iff C \ge \frac{1}{8}$. So you have a mistake somewhere.

Can anyone look for a solution that uses this kind of approach? For $n=2$ we have $(x_1+x_2)^4=({x_1}^2+{x_2}^2+2x_1x_2)^2 \ge 2\sqrt{({x_1}^2+{x_2}^2)2x_12x_2}^2=8x_1x_2({x_1}^2+{x_2}^2)$,so $C=\frac{1}{8}$ in this case.Now induct on $n$.

I did it with Lagrange multipliers. The problem is equivalent to maximizing $f(x_1,...,x_n) = \displaystyle\sum_{i<j} x_i x_j(x_i^2 + x_j^2) = \sum_i (1-x_i)x_i^3$, with the constraint that $\sum x_i = 1$. Using the Lagrangian, at a maximum that isn't on the boundary $ 4x_i^3 - 3x_i^2 = 4x_j^3 - 3x_j^2\hspace {10pt} \forall i,j$. $f(\dfrac 1 n, ..., \dfrac 1 n) = \dfrac {n-1} {n^3} < f(\dfrac 1 2,\dfrac 1 2, 0, ..., 0)$ for $n \geq 3$. The function $g(x) = 4x^3 - 3x^2$ has a local maximum at 0 and a local minimum at $x=\dfrac 1 2$. So the only other possibility for a maximum is a point of the type $(w,z,...,z)$ where $w > \dfrac 1 2, 1-w = (n-1)z$. Let $n_1 = n-1$. In that case, $f(w,z,...,z) = n_1(1-z)z^3 + n_1z(1-n_1z)^3$. But $n_1(1-z)z^3 < (1-n_1z)(n_1z)^3$ for $n_1 \geq 2$, since $z < \dfrac 1 {2n_1}$. So $f(w,z,...,z) < f(w,1-w)$. But for $n=2$, at the maximum $4((1-w)^2+(1-w)w + w^2) = 3((1-w)+w)= 3$, which is only true if $w=\dfrac 1 2$. So the max possible value of $f$ for any $n$ is $\dfrac 1 8$, at a point of the type $(\dfrac 1 2, \dfrac 1 2, 0, ..., 0)$.

orl wrote: Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such the inequality \[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C \left(\sum_{i}x_{i} \right)^4\] holds for any $x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality. Answer. $C=\tfrac{1}{8}$. (Proof) For equality, let $x_1=x_2=1, x_3=\dots=x_n=0$. Let us induct on $n \ge 2$. Now, for $n=2$ it is clear that $8ab(a^2+b^2) \le (a+b)^4$ for all $a, b \ge 0$. For $n>2$, pick $x_1, x_2$ to be the smallest two numbers in the set $\{x_1, \dots, x_n\}$. Fix $x_1+\dots+x_n=1$. Then we have $x_1+x_2 \le \tfrac{2}{n} \le \tfrac{3}{4}$ for all $n>2$, so $$f(0, x_1+x_2, \dots, x_n)-f(x_1, x_2, \dots, x_n)=\left(3\sum_{k>2} x_k\right)x_1x_2(x_1+x_2)-x_1x_2(x_1^2+x_2^2)=x_1x_2(3-4(x_1+x_2)) \ge 0$$hence we have reduced to the case of $n-1$ variables and induction applies. $\blacksquare$

Well at least I had fun while doing this,I started as $$\left (\sum_{i=1}^n x_i\right)^4=\left(\sum_{i=1}^n x_i^2+2\sum_{i<j} x_ix_j\right)^2\leq 8 \left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i>j} x_ix_j \right)$$The rest is completely obvious,right?Well wrong ,i got distracted and went completely blank and discarded the whole thing altogether. Now if we fix $\sum_{i=1}^n x_i$ we have to maximize the symmetric LHS and smoothing could really be magical here.Also note that : $$x_1x_i(x_i^2+x_1^2)+x_2x_i(x_i^2+x_2)=x_i(x_1+x_2)(x_i^2+(x_1+x_2)^2-3x_1x_2)$$The only summand that is left is $x_1x_2(x_1^2+x_2^2)$ but we can take care of him by making the $x_1,x_2$ two smallest numbers and then $x_ix_1x_2(x_1+x_2)$ would kill it easily,hence we just showed $F(x_1,x_2,...x_n)\leq F(x_1+x_2,x_3,...x_n)$ awesome right?The only case left is where there's nothing to take on the $x_1x_2(x^2+x_2^2)$ which is just the case $n=2$ and for that we easily have $C=\tfrac{1}{8}$.Note that we should clean out all the zeroes right away just to avoid some mess with equality cases. So after reading the official solution i see one is smoothing and the other one is what i did at the beginning.

From setting $x_1=x_2$ in $n=2$, we get $C\ge 1/8$. Now we prove $C=1/8$ works. Now suppose $n>2$ and let $x_1\ge ...\ge x_n>0$ (note if $x_n=0$ then it reduces to the $n-1$ case). Let $S=\sum_{i<j} x_ix_j(x_i^2+x_j^2)$, and note that if we replace $(x_{n-1}, x_n)$ with $(x_{n-1}+x_n, 0)$, then $S$ increases by $(\sum_{i=1}^{n-2} 3x_{n-1}x_n(x_{n-1}+x_n)x_i))-x_{n-1}x_n(x_{n-1}^2+x_n^2)$, which is at least $0$ for $n>3$. Thus, this replacement makes $S$ increase, meaning we've reduced the problem to the $n=2$ case. But the $n=2$ case reduces to $(x_1-x_2)^4\ge 0$, so hence we are done.

The answer is $C =\frac{1}{8}$ and equality occurs when two variables are equal and the rest are $0$. Since $(a+b)^2 \ge 4ab$, we have \begin{align*} \left(\sum_{i=1}^n x_i \right)^4 &= \left( \sum_{i=1}^n x_i^2 + 2\sum_{i <j } x_ix_j \right)^2 \\ &\ge \left( 4 \sum_{i=1}^n x_i^2 \right) \left( 2 \sum_{i <j } x_ix_j \right) \\ &= 8 \sum_{k=1}^n x_k^2 \sum_{i<j} x_ix_j \\ &\ge 8 \sum_{i<j} (x_i^2+x_j^2) x_i x_j \end{align*}where the last inequality only occurs when $n-2$ of the variables are equal to $0$. In the case when $x_1 = x_2$ and $x_3 = \cdots = x_n = 0$, the inequality reduces to $(x_1-x_2)^4 \ge 0$.

The least such constant is $C=\frac{1}{8}$. By taking $n=2,x_1=x_2$, it is clear that $C\ge \frac{1}{8}$, hence it remains to show that $C=\frac{1}{8}$ always works. By homogeneity, we may assume $\displaystyle\sum_{i=1}^n x_i=1$, hence we wish to show that \[\sum_{1\le i<j\le n}x_ix_j(x_i^2+x_j^2) \le \frac{1}{8}\]given that the sum of the $x_i$ is $1$. WLOG assume $x_1\ge x_2\ge \cdots \ge x_n$. We claim that if $n>2$, then we can increase that sum by replacing $x_1$ with $x_1+x_n$ and $x_n$ with $0$ as long as $x_1+x_n<3/4$. It is sufficient (and equivalent) to show \[\sum_{1<i<n}\left(x_1x_i\left(x_1^2+x_i^2\right)+x_nx_i\left(x_n^2+x_i^2\right)\right)+x_1x_n\left(x_1^2+x_n^2\right) < \sum_{1<i<n}(x_1+x_n)x_i\left((x_1+x_n)^2+x_i^2\right)).\]We explicitly compare the expressions in terms of $x_i$ to get that they have difference \[(x_1+x_n)\left((x_1+x_n)^2+x_i^2\right))-\left(x_1\left(x_1^2+x_i^2\right)+x_n\left(x_n^2+x_i^2\right)\right)=\]\[x_1^3+3x_1^2x_n+3x_1x_n^2+x_n^3+x_1x_i^2+x_nx_i^2-x_1^3-x_1x_i^2-x_n^3-x_nx_i^2 = 3x_1x_n(x_1+x_n)\]times $x_i$. Then, it is sufficient to show \[x_1x_n(x_1^2+x_n^2) < 3x_1x_n(x_1+x_n)\sum_{1<i<n}x_i.\]Equivalently, it is sufficient to show \[x_1^2+x_n^2 < 3(x_1+x_n)\sum_{1<i<n}x_i = 3(x_1+x_n)(1-x_1-x_n).\]In fact, taking $x_1+x_n=a$ yields \[3a(1-a)-x_1^2-x_n^2 \ge 3a-3a^2-a^2 = (3-4a)a > 0,\]as desired. Then, we can continue iterating this "merging" operation until the sum of any two nonzero $x_i$ is at least $3/4$. Remark that if we had three nonzero $x_i$ at some point, the sum of the two minimal such $x_i$ would be at most $2/3$, hence there are only two nonzero $x_i$. Now, it suffices to show the inequality \[x_1x_2(x_1^2+x_2^2) \le 1/8\]for any nonnegative $x_1,x_2$ with sum $1$. Dehomogenize to see that it suffices to show the inequality \[8a^3b+8ab^3 \le (a+b)^4.\]However, this inequality is simply a rearrangement of the statement $(a-b)^4\ge 0$, which is trivially true. Hence, equality holds at \[(x_1,\dots , x_n) = (1/2,1/2,0,\dots , 0)\]and rearrangements thereof.

I claim that the answer is $C = \frac{1}{8}$ for all $n \geq 1$. Indeed, we can check that equality is achieved when $x_1 = x_2$ and $x_3 = \ldots = x_n = 0$ for all $n$. There is nothing to prove for $n = 1$, and when $n = 2$, it is clear that $8ab(a^2 + b^2) \leq (a + b)^4$ for all nonnegative reals $a, b$ since\[8ab(a^2 + b^2) \leq 8ab(a^2 + b^2) + (a - b)^4 = (a + b)^4.\]When $n \geq 3$, we let $x_1 + x_2 + \ldots + x_n = 1$ because the inequality is homogeneous. Suppose we rearrange the numbers $x_1 \leq x_2 \leq \ldots \leq x_n$ into nondecreasing order since LHS and RHS are both symmetric. We finish this problem with a bit of smoothing: note that if we replace $(x_1, x_2)$ with $(0, x_1 + x_2)$ then the LHS changes by $x_1x_2(3(x_1 + x_2)(x_3 + \ldots + x_n) - (x_1^2 + x_2^2)$ which can be obtained by a bit of expanding. If we let $x_1 + x_2 = t$, then this expression becomes\[x_1x_2(3t(1-t) - t^2 + 2x_1x_2) = x_1x_2(t(3 - 4t) + 2x_1x_2) \geq 0\]since $x_1 + x_2 \leq \frac{2}{n} \leq \frac{2}{3}$ for all $n \geq 3$. Hence, this "smooth" does indeed increase (or rather, nondecresase) the LHS. We have reduced this to a $n-1$ variable inequality and then we can just induct down to $n = 2$ and get our desired result for all $n \geq 3$. $\blacksquare$

The answer is $\boxed{C = \frac{1}{8}}$ with equality holding at $(x_1,x_2,\dots,x_n) = (a,a,0,\dots, 0)$ for some non-negative $a$, and permutations. Claim: [Two-variable version] $x_1x_2(x_1^2+x_2^2) \leq \frac{1}{8} (x_1+x_2)^4.$ Proof: Holds due to the expansion of $(x_1-x_2)^4 \geq 0$. Claim: [Smoothing] Suppose $n \geq 3$ and for each $i$ let $0 \leq x_i \leq x_{i+1}$. Replacing $(x_1,x_2)$ with $(0,x_1+x_2)$ never decreases the left-hand side of the inequality. Proof: Check this by some algebra. Repeating smoothing the variables of our inequality, we may reduce the problem to the version where only $2$ of the variables are non-negative, which has already been proven. $\square$

We claim that the answer is $\boxed{C=\frac{1}{8}}$ for all $n$, and the equality case occurs when $x_1=x_2>0$ and $x_i=0, i>2$ or permutations, which clearly works. We prove that $C=\frac{1}{8}$ works, and that there are no other equality cases. Because the desired inequality is homogenous, WLOG let $x_1+\ldots+x_n=1$. Notice that \begin{align*} \sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) &= \sum_{i<j} \left(x_i^3x_j+x_j^3x_i \right) \\ &= \sum_{i<j} i^3j +\sum_{i>j} i^3j \\ &= \sum_{i=1}^n x_i^3(x_1+\ldots+x_{i-1}+x_{i+1}+\ldots+x_n) \\ &= \sum_{i=1}^n x_i^3(1-x_i). \end{align*}We want to prove that this expression is less than $C=\frac{1}{8}$, because we set $\sum_{i}x_{i}=1$. Claim 1: Any optimal construction can have at most one number $x_i$ in the range $(0,0.5)$. Proof. Consider the function $f(x)=x^3(1-x)$. It is easy to see that $f''(x)=6x(1-2x)$, thus making the function convex on $(0,0.5)$. Now consider the following operation: Operation: Given two indices $i,j$ so that $x_i, x_j \in (0,0.5)$ and $x_i\le x_j$, decrease $x_i$ by a small amount $\epsilon$, and increase $x_j$ by the same amount, making sure that $x_i\ge 0, x_j\le 0.5$. This configuration is higher than the previous configuration because $f(x)$ is convex on $(0,0.5)$. Therefore, an optimal construction cannot be operated on; thus it can only have at most value in $(0,0.5)$. $\blacksquare$ Claim 2: The optimal configuration contains at most two nonzero elements. Proof. Notice that $x_i=x_j=0.5$ is the only way to have more than one term at least $0.5$, and this contains two nonzero terms. This means that any other optimal construction could only have one value at least $0.5$. Because it can have at most one value in $(0,0.5)$, such a construction would have at most two nonzero values, proving the claim. $\blacksquare$ Therefore we are reduced to proving $x_ix_j(x_i^2+x_j^2)\le \frac{1}{8}$ given $x_i+x_j=1$. Notice that $x_i^2+x_j^2=(x_i+x_j)^2-2x_ix_j=1-2x_ix_j$, so letting $a=x_ix_j$ it suffices to prove that $$a(1-2a)\le \frac{1}{8}\Rightarrow -2\left(a-\frac{1}{4}\right)^2\le 0,$$which is clearly true with equality only when $a=\frac{1}{4}$. Thus we have proven that $C=\frac{1}{8}$ works. Combined with $x_i+x_j=1$, this implies $x_i=x_j=\frac{1}{2}$. Taking off the condition that the sum of the $x_i$ is $1$, we get the complete equality case to be the one at the beginning of the proof. The proof is complete.

ISL marabot solve We claim the answer is $\boxed{\frac{1}{8}}$. Part 1: Prove that the inequality always holds for $C=\frac{1}{8}$. We have \[\left(\sum_i x_i\right)^4=\left(\sum_i x_i^2+2\sum_{i<j}x_i x_j\right)^2\ge 8\left(\sum_i x_i^2\right)\left(\sum_{i<j} x_ix_j\right)=8\sum_{i<j} \left(x_ix_j (x_1^2+x_2^2+\ldots+x_n^2)\right)\ge 8\sum_{i<j}x_ix_j(x_i^2+x_j^2),\]which proves the inequality always holds for $C=\frac{1}{8}$ $\blacksquare$. Part 2: Characterize the instances of equality. Note for equality to hold, we need \[\sum_{i<j}(x_ix_j(x_1^2+x_2^2+\ldots+x_n^2))=\sum_{i<j}(x_ix_j)(x_i^2+x_j^2).\] Case 1: There exist $i$ and $j$ with $i<j$ so that $x_i, x_j>0$. Note $x_ix_j(x_1^2+x_2^2+\ldots+x_n^2)=x_ix_j(x_i^2+x_j^2)$, so all other terms of the sequence are $0$. We have \[x_ix_j(x_i^2+x_j^2)=\frac{1}{8}(x_i+x_j)^4\implies 8x_i^3x_j+8x_j^3x_i=x_i^4+4x_i^3x_j+6x_i^2x_j^2+4x_ix_j^3+x_j^4\implies x_i^4-4x_i^3x_j+6x_i^2x_j^2-4x_ix_j^3+x_j^4=(x_i-x_j)^4=0\implies x_i=x_j\] Case 2: There exists a unique $i$ so that $x_i>0$. It's easy to see this doesn't satisfy equality in the original inequality (LHS is $0$ and RHS is greater than $0$). Case 3: $x_i=0$ for all $i$. This works. So the instances of equality are $\boxed{x_i=x_j>0\text{ for some } i<j\text{ and everything else in the sequence is }0\text{ or }x_i=0\forall i}$ $\blacksquare$ Now we will show that no lower $C$ works. Checking the first equality instance gives \[2x_i^4\le16Cx_i^4\implies 2\le16C\implies \frac{1}{8}\le C\]$\blacksquare$

We claim that the minimum value of $C$ is $\frac18$. Substituting $x_1=x_2=1$ and $x_3=x_4=\ldots=x_n=0$ gives $2\leq16C$, so $C\geq\frac18$. Now, we will prove that $C=\frac18$ works. Suppose that there are at least three nonzero $x_i$. Then, assume without loss of generality $x_1\geq x_2\geq x_3>0$. We claim that replacing $x_2$ with $x_2+x_3$ and $x_3$ with $0$ will increase the left hand side. We have \begin{align*} &x_2x_3(x_2^2+x_3^2)+(x_2^3+x_3^3)(x_1+x_4+\ldots+x_n)+(x_2+x_3)(x_1^3+x_4^3+\ldots+x_n^3)\\ &-(x_2+x_3)^3(x_1+x_4+\ldots+x_n)-(x_2+x_3)(x_1^3+x_4^3+\ldots+x_n^3)\\ =&x_2^3x_3+x_2x_3^3+(x_1+x_4+\ldots+x_n)(x_2^3+x_3^3-(x_2+x_3)^3)\\ =&x_2^3x_3+x_2x_3^3-3x_2x_3(x_2+x_3)(x_1+x_4+\ldots+x_n))\\ \leq&x_2^3x_3+x_2x_3^3-3x_1x_2x_3(x_2+x_3)\\ \leq&x_2x_3(x_2^2+x_3^2-(x_2+x_3)^2)\\ =&x_2x_3(-2x_2x_3)\\ <&0. \end{align*}Therefore, we only need to show $C=\frac18$ works when at most two of the $x_i$ are nonzero. This is equivalent to showing $x_1^4+4x_1^3x_2+6x_1^2x_2^2+4x_1x_2^3+x_2^4\geq8x_1^3x_2+8x_1x_2^3$, which is equivalent to $(x_1-x_2)^4\geq0$. Therefore, $C=\boxed{\frac18}$.

The answer is $C=\frac{1}{8}$. Equality holds when two of the values are equal and the rest are equal to $0$. The inequality is homogeneous, so WLOG set $\sum x_i=1$. Then we wish to find the least constant $C$ such that the inequality $\sum (x_i^3-x_i^4)\le C$ holds for all nonnegative $x_i$, or equivalently, maximize the value of $\sum (x_i^3-x_i^4)$. Now suppose that the answer is $C=\frac{1}{8}$ for all $2\le i\le n$. We show that the answer will be $\frac{1}{8}$ for $n+1$ (we will deal with the base case at the end of the proof). Clearly when $n$ or $n+1$ of the variables are zero, the whole sum becomes zero, which is not optimal. Otherwise, if we have anywhere from $1$ to $n-1$ zeros, the maximum value is $\frac{1}{8}$ by the inductive hypothesis. It suffices to show, then, that when none of the values are zero, we have $\sum (x_i^3-x_i^4)<\frac{1}{8}$. It would be enough to show that there exist $x_i$ and $x_j$ that may be combined into $x_i+x_j$ such that the new sum (which has $n$ variables, and is therefore at most $\frac{1}{8}$) is greater than the old sum. Then we require \[(x_i+x_j)^3-(x_i+x_j)^4>x_i^3-x_i^4+x_j^3+x_j^4\]which simplifies to \[3x_i+3x_j>4x_i^2+6x_ix_j+4x_j^2\]so it would be enough that \[3x_i+3x_j\ge 4(x_i+x_j)^2>4x_i^2+6x_ix_j+4x_j^2\implies x_i+x_j\le \frac{3}{4}.\]Suppose that such $x_i$ and $x_j$ do not exist. Then for every pair of distinct $i,j$ we must have $x_i+x_j>\frac{3}{4}$ which implies that every $x_i$ is less than $\frac{1}{4}$, a contradiction. Therefore we may keep reducing the number of terms, which easily gives that the maximum occurs when $n-1$ of the variables are equal to zero while the other two are nonzero and sum to $1$. Now, we wish to maximize $x^3-x^4+y^3-y^4$ subject to the condition $x+y=1$. This is equal to \[-2x^4+4x^3-3x^2+x.\]The derivative of this is equal to $(1-2x)^3$ which means that we need only to test $x=0,\frac{1}{2},1$. Clearly $x=\frac{1}{2}$ is optimal, which implies $y=\frac{1}{2}$ and gives $C=\frac{1}{8}$. Homogenizing, we find that equality occurs when two variables are the same and the rest are equal to $0$. We are done. $\blacksquare$

Very nice solution

Let $x_1=x_2$ to get $2\le 16C$ so $C\ge \frac{1}{8}.$ We claim that $C=\frac{1}{8}$ is sufficient. We homogenize $x_1+x_2+\dots+x_n=1.$ The desired inequality becomes \[\sum_{i=1}^{n}{x_i^3(1-x_i)}\le \frac{1}{8}.\]Since $f(x)=x_i^3(1-x_i)$ is convex in $\left(0,\tfrac{1}{2}\right)$, we can apply Karamata's inequality: \[f(x_1)+f(1-x_1)+(n-2)f(0)\ge f(x_1)+f(x_2)\dots+f(x_n)\]which is the desired. This requires $(0.5,0.5,0,\dots)$, up to scaling or reordering.

Note that $(1,1,0,0, \dots , 0)$ gives that $C \ge \frac{1}{8}$. We prove that $C=\frac{1}{8}$ works. Note that $$\frac{1}{8}\left( \sum_{i}x_i \right) ^4 = \frac{1}{8} \left( \sum_{i}x_i^2+2 \sum_{i<j} x_ix_j \right)^2 \ge \sum_{i}x_i^2 \cdot \sum_{i<j} x_ix_j = \sum_{i<j} x_ix_j\left(\sum_{k}x_k^2\right) \ge \sum_{i<j} x_ix_j(x_i^2+x_j^2).$$For equality to hold we need $x_ix_j\left(\sum_{k}x_k^2\right)=x_ix_j(x_i^2+x_j^2)$. This is obviously false if there is more than $2$ nonzero $x_i$'s by taking $x_i$ and $x_j$ to be nonzero. If there is at most one nonzero $x_i$ we have that $(0,0, \dots 0)$ works. If there is $2$ nonzero $x_i$'s then we get the solution $(a,a,0,0, \dots , 0)$ and permutations.

Excellent

We claim that the answer is $\frac{1}{8}$. By taking $x_1=x_2=1/2$ and everything else 0, we can see that no smaller $C$ works. We will then show that 1/8 works. Since the inequality is homogenous, we can WLOG that $x_1+x_2\cdots + x_n=1$. Then, the inequality becomes $$\sum_{i=1}^n x_i^3(1-x_i)\leq \frac{1}{8}.$$ Note that $$f(x)=x^3-x^4$$is convex in the interval $[0,1/2].$ We claim that if $$\sum_{i=1}^n x_i^3(1-x_i)$$is maximized, then we cannot have 3 or more nonzero variables. Suppose FTSOC that the sum is maximized while having three nonzero variables. Then, we can find two variables that are both strictly between 0 and 1/2, since there are three nonzero variables. However, since $f$ is convex on $[0,1/2]$, we can "push" those two chosen variables away from each other until one of them reaches 0 or 1/2. Due to Karamata's, this pushing increases the value of the sum, contradiction. Therefore, we can reduce this to 2 variables instead of 4, in which case it is easy to see that both being $1/2$ is optimal, giving the equality case. Undoing the homogenization, equality occurs when two variables are equal and the rest are all 0.

I tried solving the problem by simplifying the LHS. This gave the following problem: Find the minimum value of C such that f(x1)+f(x2)+.....f(xN) <= c for all integers N>1 and positive reals x1, x2, .......xN summing to 1, where f(x) = x^3-x^4. I figured out that f is convex from x=0 to x=1/2, and used Jensen's inequality to figure out that, if x1,x2,.....xN are all at most 1/2, then the LHS is at least Nf(1/N), and if N=1/2, then Nf(1/N) = 1/8. I'm trying to find a maximum value but this only leads to a set of minimum values. Did i use the wrong method or is there a way to solve this with Jensen's?

We claim $C=1/8$. Let $x_1=x_2 >0$ and $x_3=x_4=\dots=x_n=0$ to get $C\geq \frac 18$. The inequality is homogenous, so let the sum of the variables be $1$. It remains to show $\sum x_i^3-x_i^4 \leq \frac 18$, but this follows by Karamata, as $x^3-x^4$ is convex over the interval $\left( 0, \frac 12\right)$ and $\left(x_1, 1-x_1, 0, \dots, 0 \right)$ majorizes $(x_1, x_2, \dots, x_n)$. Equality is when $x_1=x_2 >0$ and the rest of the variables are $0$.

There is an SOS expression. \[\left(\sum_{i=1}^na_i\right)^4-8\sum_{1\le i<j\le n}a_ia_j\left(a_i^2+a_j^2\right)=\left(\sum_{i=1}^na_i^2-2\sum_{1\le i<j\le n}a_ia_j\right)^2+8\sum_{\substack{1\le k\le n,\\1\le i<j\le n,\\k\ne i,j}}a_k^2a_ia_j.\]

Answer is $C = \frac 18$. The idea is to do smoothing (or Karamata if you like high-tech stuff.) Fix $\sum_{i=1}^n x_i = 1$, and consider the change $(x_1, x_2) \to (0, x_1+x_2)$. This alters the LHS by \begin{align*} A &= x_1 \sum_{j=2}^n x_j(x_1^2+x_j^2) + x_2\sum_{j=3}^n (x_2^2+x_j^2) - (x_1+x_2) \sum_{j=3}^n x_j((x_1+x_2)^2+x_j^2) \\ &= 3(1-x_1-x_2)(x_1+x_2) - x_1^2-x_2^2. \end{align*}One can check that this is strictly greater than zero for $x_1 + x_2 < 1$. Thus we can reduce the problem to the case where $x_1 = x_2 = k$ and all the rest of the $x_i$ are $0$. This yields $C = \frac 18$.

Wow, this was very technically involved. We claim the minimum is achieved at $C = \frac{1}{8}$, with equality at $x_1 = x_2 = \frac{1}{2}$ and $x_3 = x_4 = \dots = x_n = 0$. Homogenize such that $\sum x_i = 1$. Then we wish to find the least constant $C$ such that, \begin{align*} \sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2) \leq C \end{align*}However the left hand side is simply, \begin{align*} \sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2) &= \sum_{i} x_i^3(1 - x_i)\\ &= \sum_i x_i^3 - \sum_i x_i^4. \end{align*}Thus we wish to find the least constant $C$ such that, \begin{align*} \sum_i (x_i^3 -x_i^4) \leq C \end{align*}It suffices to maximize the left hand side. Note that the function $f(x) = x^3 - x^4$ is convex on the interval $\left(0, \frac{1}{2}\right)$ and that there is at most one $x_i > \frac{1}{2}$, which we assume wlog is $x_1$. Now Karmata's gives, \begin{align*} f(x_1) + f(1- x_1) + (n-2) \cdot f(0) &\geq \sum f(x_i)\\ f(x_1) + f(1- x_1) &\geq \sum f(x_i) \end{align*}Thus we wish to maximize, \begin{align*} x^3 - x^4 + (1-x)^3 + (1-x)^4. \end{align*}However the derivative of this is simply $(1-2x)^3$ from which we finish, with equality achieved at $x = \frac{1}{2}$.

I claim that the answer is $C = \frac{1}{8}$. Note that $n = 2$ and $x_1 = x_2 = 1$ implies $C \geq \frac{1}{8}$. To show $C = \frac{1}{8}$ works, just note \[ \left( \sum x_i\right)^4 = \left( \sum x_i^2 + 2\sum x_ix_j \right)^2 \stackrel{AM-GM}{\ge} \left( 2\sqrt{2} \left( \sum x_i^2\right) \left( \sum x_ix_j\right) \right)^2 = 8 \left( \sum x_i^2\right) \left( \sum x_ix_j \right) \]One can easily check that the last expression is at least $8 \sum (x_i^2 + x_j^2)x_ix_j$. Note that equality holds only if \[ \sum x_i^2x_jx_k = 0 \]which implies that all $x_i$ are zero except for say $x_1$ and $x_2$. But then $x_1^2 + x_2^2 = 2x_1x_2$ implies that $x_1 = x_2$. Since $x_1 = x_2$ and $x_3 = \dots = x_n = 0$ also implies equality, this is the only equality case. $\blacksquare$

$\left(\sum x_i \right)^4 = \left(\sum x_i^2 + 2\sum x_ix_j \right)^2 \geq 8\left( \sum x_i^2 \right)\left(\sum x_ix_j \right)$ $= 8\left(x_jx_k \left(\sum x_i^2\right)\right) \geq \sum x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right)$, equality holds iff $n-2$ of them are $0$ and other $2$ are equal.

We claim that $\min(C)=\frac{1}{8}$. Plugging in $x_1=x_2$ for $n=2$ yields that $C \geq \frac{1}{8}$ so we cannot do better. Now, we try to prove the following inequality, \[\sum_{1\leq i < j \leq n}x_ix_j(x_i^2+x_j^2)\leq \frac{1}{8} \left(\sum_{1\leq i \leq n}x_i\right)^4\]We immediately observe that the given inequality is homogeneous and thus, assume WLOG $x_1+\dots +x_n=1$. Then, we smooth this inequality. Assume $0<x_1\leq x_2 \leq \dots \leq x_n$. Consider smoothing as follows. We simply do the transformation $(x_1,x_2) \mapsto (0,x_1+x_2)$. This changes the LHS by, \[\sum_{i=3}^n x_i(x_1+x_2)((x_1+x_2)^2 + x_i^2) - \sum_{i=3}^n x_1x_i(x_1^2+x_i^2) - \sum_{i=3}^n x_2x_i(x_2^2+x_i^2) - x_1x_2(x_1^2+x_2^2)\]Simplifying this expressions yields, \begin{align*} S &= (x_1+x_2)^3(1-x_1-x_2)-(x_1^3+x_2^3)(1-x_1-x_2)-x_1x_2(x_1^2+x_2^2)\\ &= x_1x_2\left(3(x_1+x_2)(1-x_1-x_2)-x_1^2-x_2^2\right) \end{align*}Now, we wish to show that this expression is positive for $x_1+x_2\leq 1$. Simply note that, \[S= x_1x_2((x_1+x_2)(3-4(x_1+x_2) +2x_1x_2)\]Thus, it is easy to see that $S>0$ when $x_1+x_2\geq \frac{3}{4}$. But since $x_1,x_2$ are the minimal terms, we know that $x_1+x_2 \leq \frac{2}{3}<\frac{3}{4}$. So $S$ is indeed positive as desired. Now, we can ignore the newly created terms of value zero and are reduced to the $n-1$ case. Thus, we are reduced to the $n=2$ case via induction. Here, we smooth again. Let $\epsilon=\frac{x_2-x_1}{2}$. We simply do the transformation $(x_1,x_2)\mapsto (x_1+ \epsilon, x_2-\epsilon)$. Then, it is easy to see that \[(x_1+\epsilon)(x_2-\epsilon)((x_1+\epsilon)^2+(x_2+\epsilon)^2) > x_1x_2(x_1^2+x_2^2)\]when $\epsilon>0$. Thus, we are left with $x_1=x_2$ at which point the given inequality is obvious. Thus, the desired inequality holds for all non-negative reals $x_1,x_2,\dots,x_n$ for any natural number $n\geq 2$, with equality holding if and only if $x_1=x_2=k$ for some $k\in \mathbb{R}_{\geq 0}$ and $x_3=x_4=\dots =x_n=0$.

The answer is $C = \frac{1}{8}.$ The cases of equality are when two of the variables are equal, and all of the others are $0.$ We have that $$\left(\sum_{i=1}^{n}x_{i} \right)^{4} $$$$= \left(\sum_{i=1}^{n}x_{i}^{2}+\sum_{1 \le i < j \le n} 2x_{i}x_{j} \right)^{2} $$$$\ge 8\left(\sum_{i=1}^{n}x_{i}^{2} \right)\left(\sum_{1 \le i < j \le n}^{n}2x_{i}x_{j} \right) $$$$= 8\left(\sum_{1 \le i < j \le n} x_{i}x_{j}(x_{i}^{2}+x_{j}^{2})+\sum_{1 \le i < j \neq k \le n} x_{i}x_{j}x_{k}^{2} \right) $$$$ \ge 8\left(\sum_{1 \le i < j \le n} x_{i}x_{j}(x_{i}^{2}+x_{j}^{2})\right). $$Now, equality holds if and only if the last expression is zero. This implies that all but at most $2$ of the variables can be nonzero. Now, by the equality of the AM-GM, we find the result.

I claim the answer is $\frac 18$. Necessity: Consider $x_1 = x_2 = 1, x_i = 0$ otherwise, the left side is just $2$, right side is just $2^4$, so $C \ge \frac 18$. Sufficiency: Smoothing. We show that we can reduce the $x_i$ into a position where exactly two of them are nonzero. Fix the sum of the $x_i$ as $s$. We show if $x_i + x_j \le \frac s2$. we can send $x_i$ to $0$ and $x_j$ to $x_i + x_j$ while strictly increasing the left side of the inequality. First observe the left side is $\sum_{sym} x_i^3x_j$, so we can write it as $\sum (s - x_i )x_i^3$. Thus we just need to prove for the $x_i = a, x_j = b$ satisfying the conditions that the inequality $ s(a + b)^3 - (a + b)^4 \ge s(a^3 + b^3) - a^4 -b^4$. Rearranging, we desire $3sab(a +b ) \ge ab(4a^2 + 6ab + 4b^2)$, dividing $ab$ we equivalently desire $3s(a + b) \ge 4a^2 + 6ab +4b^2$, since $s \ge 2(a + b)$ this is equivalent to proving $6(a + b)^2 \ge 4a^2 + 6ab + 4b^2$, which is obviously true. Now repeating this operation a finite number of times forces us into a position where only two $x_i$ are nonzero and the left hand side of the inequality is greater than before while the right hand side is unchanged, so it suffices to only analyze these cases. We just want to prove then that $a^3b + b^3a \le \frac 18 (a + b)^4$, or that $0 \le (a - b)^4$, which is trivial.

The answer is $C = \boxed{\tfrac{1}{8}}$, with equality when $x_1=x_2 = \tfrac{1}{2}$ and all other $x_i=0$. WLOG assume that $x_1 \le x_2 \le \dots \le x_n$, and homogenize the equation such that $x_1+x_2+\dots +x_n = 1$. We will proceed by smoothing; consider the change $(x_1, x_2) \to (0,x_1+x_2)$. This alters the LHS by \begin{align*} S &= x_1 \sum_{i=2}^n x_i(x_1^2+x_i^2) + x_2\sum_{i=3}^n x_i(x_2^2+x_i^2) - (x_1+x_2) \sum_{i=3}^n x_i((x_1+x_2)^2+x_i^2) \\ &= x_1x_2(x_1^2+x_2^2) + x_1 \sum_{i=3}^n (x_i(-2x_1x_2-x_2^2)) + x_2 \sum_{i=3}^n (x_i(-2x_1x_2-x_1^2)) \\ &= x_1x_2(x_1^2+x_2^2) - (3x_1^2x_2+3x_1x_2^2) \cdot \sum_{i=3}^n x_i \\ &= x_1x_2(x_1^2+x_2^2) - 3x_1x_2(x_1+x_2) \cdot \sum_{i=3}^n x_i. \end{align*} Then, letting $x_1+x_2=k$, we can simplify this further to \begin{align*} S &= x_1x_2(x_1^2+x_2^2) - 3kx_1x_2 \cdot (1-k) \\ &= x_1x_2(x_1^2+x_2^2 - 3k(1-k)) \\ &= x_1x_2(k^2-2x_1x_2-3k(1-k)) \\ &= -x_1x_2(k(3-4k)+2x_1x_2). \\ \end{align*} Since $k = x_1+x_2 \le \frac{2}{n} \le \frac{2}{3}$, we have $k(3-4k)>0$, which means that $S<0$. Hence, the LHS is increasing, so we may smooth all but two of the variables to $0$. Suppose that $x_1$ and $x_2$ are the only two nonzero variables. We wish for the following inequality to hold: \[x_1x_2(x_1^2+x_2^2) \le C (x_1+x_2)^4.\] Noting that $x_1+x_2=1$, we can simplify this to \[x_1x_2(1-2x_1x_2) \le C.\] At this point, we may write the LHS as a quadratic in terms of $x_1x_2$ to solve that $C \ge \tfrac{1}{8}$.

Note that taking $x_1=x_2=1/2$ and the rest $0$ it follows that $C \geq \frac 18$. Thus, in order to prove that $C=\frac 18$, it suffices to show $$\frac 18 \left (\sum_{i=1}^n x_i\right)^4\geq \sum_{i=1}^{n-1}\sum_{j=i+1}^n x_i x_j(x_i^2+x_j^2) \qquad (1)$$for all $n\geq 2$. Note that $$\frac 18 (x+y)^4-xy(x^2+y^2)=\frac 18 (x-y)^4\geq 0.$$This proves $(1)$ for $n=2$. Let us proceed with induction. Suppose that we have proven $(1)$ for some $k\geq 2$. Now we will prove it for $k+1$. Consider $y_1\geq y_2\geq \dots \geq y_{k+1}\geq 0$. Take $x_1=y_1$, $x_2=y_2$, $\dots$, $x_{k-1}=y_{k-1}$, $x_k=y_k+y_{k+1}$ then by calculating we get $$\sum_{i=1}^{k-1}\sum_{j=i+1}^k x_i x_j (x_i^2+x_j^2)-\sum_{i=1}^k \sum_{j=i+1}^{k+1} y_i y_j (y_i^2+y_j^2)= (3y_k y_{k+1}^2+3y_k^2 y_{k+1})\sum_{i=1}^{k-1} y_i-y_k^3 y_{k+1}-y_k y_{k+1}^3 \geq 0.$$Notice that there is no equality above if $y_{k+1}>0$. It follows that $$\sum_{i=1}^k \sum_{j=i+1}^{k+1} y_i y_j (y_i^2+y_j^2)\leq \sum_{i=1}^{k-1}\sum_{j=i+1}^k x_i x_j (x_i^2+x_j^2) \leq \frac 18 \left(\sum_{i=1}^k x_i\right )^4=\frac 18 \left(\sum_{i=1}^{k+1} y_i\right )^4.$$This proves $(1)$ for $k+1$. This finishes the proof. Also from above it's clear that the equality holds only when two of the $x_i$s are equal and rest are $0$.