according to mathematical reflections it is due to vasc majorization and am gm are keys and this is closely related to imo 99a2,hope you get ,if try.

It must be 1/16 ,when two of them are 1/2.

what is mathematical reflections?

ZHANGWENZHONGKK wrote: It must be 1/16 ,when two of them are 1/2. Obviosly, maximum when $ x_i=1,i<5, x_j=0,j\not =0$ and $ max=1$.

Rust wrote: ZHANGWENZHONGKK wrote: It must be 1/16 ,when two of them are 1/2. Obviosly, maximum when $ x_i = 1,i < 5, x_j = 0,j\not = 0$ and $ max = 1$. What?

Can anyone post a solution please?I really like this problem.

ZHANGWENZHONGKK wrote: It must be 1/16 ,when two of them are 1/2. $ (x_1,x_2) = \left(\frac{4}{5},\frac{1}{5}\right) \Rightarrow (x^4_1 - x^5_1)+(x^4_2 - x^5_2) = \frac{52}{625} = 0.0832 > \frac{1}{16} = 0.0625$.

Here is a sketch of my method. The function $ f(x)=x^4-x^5$ is convex in some intervals and concave in others. Let's say $ x_1,x_2,...,x_i$ are in the interval in which it is convex and $ x_n,x_{n-1},...,x_{i+1}$ are in the interval in which it is concave. Then it follows that the maximum occurs when $ x_n=x_{n-1}=\cdots=x_{i+1}$ and $ x_2=x_3=\cdots=x_i=0$ and $ x_1=1-(n-i)x_n$. Now we are simply maximizing $ (n-i)(x_n^4-x_n^5)+(1-(n-i)x_n)^4-(1-(n-i)x_n)^5$, which should be fairly straightforward. In the end I got the maximum when all $ x_i$ are equal.

maximum, when $ x_1=\frac{3+\sqrt 3}{6},x_2=\frac{3-\sqrt 3}{6}, x_3=...=x_n=0$, Then $ f_max=\frac{1}{12}=0.83333333...$.

Yes, sorry. My last few logic steps were mistaken. One finds that the maximum when looking at $ n-i$ occurs when $ n-i$ is maximal. Since $ \sum x_i=1$ and since $ f$ is concave on $ x\in(\frac{3}{5},\infty)$, it follows that the maximum of $ n-i$ is $ 1$. Now we have a quintic of which we need to find the maximum. The derivative turns out to be $ -(-1 + 2 x) (1 - 6 x + 6 x^2)$, and so the maximum is what Rust said.

WOW I'm literally surprised that this thread hasn't got a full solution for 13 years... The main idea is SMOOTHING. Let's follow sir CANBANKAN's technique: check small cases and then generalize. So we first check $n = 2,3$. It's easy to show that there are at most two $0$s among $x_1, \cdots, x_n$ when $\sum_{j=1}^n(x_j^4-x_j^5)$. Therefore, consider a smooth $(x,y)\to (x+y,0)$. We wish to show $$(x+y)^4-(x+y)^5>x^4-x^5+y^4-y^5\iff 4x^2+4y^2+6xy>5x^3+5y^3+10x^2y+10xy^2$$ The LHS of the above inequality $ = \frac 72(x^2+y^2)+\frac 12(x^2+y^2)+6xy\geqslant \frac 72(x^2+y^2)+xy+6xy = \frac 72(x+y)^2$, while the RHS $\leqslant 5x^3+5y^3+15x^2y+15xy^2 = 5(x+y)^3$. Henceforth if we make sure $x+y<\frac {7}{10}$ the above inequality holds. So we have the following key claim: Claim. If $x+y<\frac {7}{10}$, then $(x+y)^4-(x+y)^5>x^4-x^5+y^4-y^5$. Denote $k$ by the number of non-negative variants among $x_1, \cdots, x_n$. WLOG suppose $$x_1\geqslant x_2\geqslant \cdots\geqslant x_k\geqslant 0, x_{k+1} = x_{k+2} = \cdots = x_n =0.$$If $k\geqslant 3$, denote $$x_i' = x_i(i = 1,2,\cdots, k-2), x_{k-1}' = x_{k-1}+x_k, x_k' = x_{k+1}' = \cdots = x_n' = 0.$$Since $x_{k-1}+x_k\leqslant \frac 2n\leqslant \frac 23 <\frac {7}{10}$, by the Claim we have $\sum_{j=1}^n(x_j'^4-x_j'^5)>\sum_{j=1}^n(x_j^4-x_j^5)$. As long as $k\geqslant 3$ the above smoothing process can be done. After at most $n-2$ smoothing, we can turn all the $x_3, \cdots, x_n$ into $0$ while $S$ doesn't decrease. Denote $x_1 = a, x_2 = b$ at this very moment respectively, for convenience. Now $a+b=1$ and $$S = a^4b+ab^4 = ab(a^3+b^3) = ab(a+b)(a^2+b^2-ab) = ab(1-3ab) = \frac 13(3ab)(1-3ab)\leqslant \frac 13\cdot\frac 14 = \frac {1}{12}.\qquad\blacksquare$$ Equality holds when $a = \frac{3+\sqrt{3}}{6}, b = \frac{3-\sqrt{3}}{6}$. Edit: Changed the tags.

Creative solution

Here's another smoothing solution, I suppose. Let $P(x_1, x_2, \cdots, x_n)$ denote the sum for variables $x_1, x_2, \cdots, x_n$. Fix $x_3, x_4, \cdots, x_n$, and note that \begin{align*} x_1^4 - x_1 ^5 + x_2^4 - x_2^5 &\leq (x_1+x_2)^4 - (x_1+x_2)^5 \\ \iff 5(x_1+x_2)(x_1^2+x_1x_2+x_2^2) &\leq 4x_1^2+6x_1x_2+4x_2^2, \end{align*}which is true if $x_1 + x_2 \leq \frac 45$. So $$P(x_1, x_2, \cdots, x_n) \leq P(0, x_1+x_2, x_3, \cdots, x_n) \leq \cdots \leq P(x, y, 0, 0, 0, \cdots, 0).$$In this case, let $x = \frac 12 + t$, $x_2 = \frac 12 - t$. With some algebra we may obtain the expression equals $$-3t^4 + \frac 12 t^2 + \frac 1{16} \leq \boxed{\frac 1{12}}$$in this case. Equality obviously holds.