In order to do smoothing, we will prove the following lemma, which is the two-variable version of the problem. Lemma: If $x,y > 0$ and $xy \ge 1$ then \[ (x^2+1)(y^2+1) \le \left( \left( \frac{x+y}{2} \right)^2+1 \right)^2. \] Proof. Subtracting the two sides gives \[ \frac1{16} (x-y)^2 \left( x^2+6xy+y^2-8 \right) \]which is nonnegative. $\blacksquare$

A tricky solution is as follows: by letting $d = \frac{a+b+c}{3}$ we get \begin{align*} (a^2+1)(b^2+1)(c^2+1)(d^2+1) &\le \left( \left( \frac{a+d}{2} \right)^2+1 \right)^2 \left( \left( \frac{b+c}{2} \right)^2+1 \right)^2 \\ &\le \left( \left(\frac{a+b+c+d}{4}\right)^2 +1 \right)^4 \\ &= \left( d^2+1 \right)^4 \end{align*} (Note the equality case is $a=b=c \ge 1$, not just $a=b=c=1$.)

I think I saw this problem before.

arqady wrote: I think I saw this problem before. It was posted some days ago, but later deleted due to obvious reasons.

Tintarn wrote: arqady wrote: I think I saw this problem before. It was posted some days ago, but later deleted due to obvious reasons. But I think the post locked. And that seems interesting how someone posted before

The problem is problem 1 Algebra section SHL IMO 2016 . My solution:

mathwizard888 wrote: Let $a$, $b$, $c$ be positive real numbers such that $\min(ab,bc,ca) \ge 1$. Prove that $$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$ Call a triple $(a, b, c)$ of positive reals valid, if $\min(ab, bc, ca) \ge 1$. Claim 1. If $(a, b, c)$ is valid then so is $\left(\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}\right)$. (Proof) For any $\{x, y, z\}=\{a, b, c\}$ we have $$\left(\frac{x+y}{2}\right)\cdot \left(\frac{x+z}{2}\right) \geqslant \sqrt{xy} \cdot \sqrt{xz} \geqslant 1,$$by AM-GM, hence the claim is true. $\blacksquare$ As a corollary, note that each of $\left(\frac{x+y}{2}\right)$ is at least $1$ by AM-GM inequality, where $x, y \in \{a, b, c\}$ and $x \ne y$. Define $f(x)=\log(1+x^2)$ for all $x>0$. Note that the given inequality is equivalent to showing $$f(a)+f(b)+f(c) \leqslant 3f\left(\frac{a+b+c}{3}\right),$$for all valid triples $(a, b, c)$. Also, $\frac{2(1-x^2)}{(1+x^2)^2} \le 0$ is the second derivative of $f$, so it is concave on $[1, \infty)$. Claim 2. For all $x, y>0$ such that $xy \ge 1$ we have $$f\left(\frac{x+y}{2}\right) \geqslant \frac{f(x)+f(y)}{2}.$$(Proof) Observe that this is equivalent to $$\left(1+\left(\frac{x+y}{2}\right)^2\right)^2 \geqslant (1+x^2)(1+y^2),$$whenever $xy \ge 1$. However, this can be seen to be equivalent to $$\frac{1}{8}(x-y)^2 \cdot \left((x+y)^2+4xy-8\right) \geqslant 0,$$which is obviously true now. $\blacksquare$ Applying Claim 2, we see that $$f(a)+f(b)+f(c) \leqslant f\left(\frac{a+b}{2}\right)+f\left(\frac{b+c}{2}\right)+f\left(\frac{c+a}{2}\right) \leqslant 3f\left(\frac{a+b+c}{3}\right).$$The last inequality holds by Jensen's on $\left(\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}\right)$ as each of them is at least $1$; so we are done.

Here it is my solution. Same way we prove $(x^2+1)(y^2+1)\leq \left(\left(\frac{x+y}{2}\right)^2+1\right)^2$ for any reals $x,y$ with $xy\geq 1$. Since from condition $ab,bc,ca\geq 1$ so we have $(a^2+1)(b^2+1)\leq \left(\left(\frac{a+b}{2}\right)^2+1\right)^2$ and $(b^2+1)(c^2+1)\leq \left(\left(\frac{b+c}{2}\right)^2+1\right)^2$ and $(c^2+1)(a^2+1)\leq \left(\left(\frac{c+a}{2}\right)^2+1\right)^2$ Multiplying this three inequalities and taking root square we have $(a^2+1)(b^2+1)(c^2+1)\leq \prod_{cyc} \left(\left(\frac{a+b}{2}\right)^2+1\right)$ Similiar way we find $\prod_{cyc} \left(\left(\frac{a+b}{2}\right)^2+1\right)\leq \prod_{cyc} \left(\left(\frac{2a+b+c}{2}\right)^2+1\right)$ since from AM-GM we have $\frac{a+b}{2}\geq \sqrt{ab}\geq 1$ and $\frac{b+c}{2}\geq 1$ and $\frac{c+a}{2} \geq 1$ hence $(a^2+1)(b^2+1)(c^2+1)\leq \prod_{cyc} \left(\left(\frac{2a+b+c}{2}\right)^2+1\right)$ With induction we prove $(a^2+1)(b^2+1)(c^2+1)\leq \prod_{cyc} \left(\left(\frac{(2^n+2)a+(2^n-1)b+(2^n-1)c}{3\cdot 2^n}\right)^2+1\right)$ for $n$ even and $(a^2+1)(b^2+1)(c^2+1)\leq \prod_{cyc} \left(\left(\frac{(2^n-2)a+(2^n+1)b+(2^n+1)c}{3\cdot 2^n}\right)^2+1\right)$ for $n$ odd Taking limit $\lim_{n\rightarrow +\infty}$ we have $(a^2+1)(b^2+1)(c^2+1)\leq \left(\left(\frac{a+b+c}{3}\right)^2+1\right)^3$ Taking cube root we have $\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)}\leq \left(\frac{a+b+c}{3}\right)^2+1$ As desired.

mathwizard888 wrote: Let $a$, $b$, $c$ be positive real numbers such that $\min(ab,bc,ca) \ge 1$. Prove that $$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$ My solution,published by the professor Bogomolny on cut-the-knot . I also informed him about the other 5 proofs from here . http://www.cut-the-knot.org/m/Algebra/ProblemFrom2016IMOShortlist.shtml

arqady wrote: I think I saw this problem before. Me too, and the date on that thread is about 2 years ago and many users replied it. However, when I visited that thread again some days ago, the whole thread had disappeared.

Generalisation: Let $a_1,a_2,\ldots a_n$ be positive reals such that the product of any two distinct $a_i$ is at least 1. Prove that \[ \prod_{i=1}^{n} (a_i^2+1) \leq \left( \sum_{1\leq i<j\leq n} \frac{2a_ia_j}{n(n-1)}+1 \right)^{n-1} \left( \sum_{i=1}^{n} \frac{a_i^2}{n}+1 \right) \]

May be using an identity: $(1+a^2)(1+b^2)(1+c^2)=(ab+bc+ca-1)^2+(a+b+c-abc)^2$

Isn't it simple, AM-GM then AM-QM inequality ??

v_Enhance wrote: This can be done directly by subtraction, since $(x-y)^2$ also factors out. Could someone please elaborate on what is happening here?

Is there anything preventing us from just smoothing to the average and from that the result would be immediate?

(I answered this by PM already but posting here to save othe people time) Delray wrote: Is there anything preventing us from just smoothing to the average and from that the result would be immediate? Smoothing $x$ and $y$ together requires $x,y \ge 1$ as stated in the lemma. So if $c$ is smaller than $1$ the lemma doesn't apply. Delray wrote: Could someone please elaborate on what is happening here? Factor $((\tfrac{2x+y}{3})^2+1)^3-(x^2+1)^2(y^2+1)$ to show it is nonnegative.

Mr.Techworm wrote: Isn't it simple, AM-GM then AM-QM inequality ?? lol that's what I thought

nvm $$ $$

anantmudgal09 wrote: Define $f(x)=\log(1+x^2)$ for all $x>0$. Note that the given inequality is equivalent to showing $$f(a)+f(b)+f(c) \leqslant 3f\left(\frac{a+b+c}{3}\right),$$for all valid triples $(a, b, c)$. Also, $\frac{2(1-x^2)}{(1+x^2)^2} \le 0$ is the second derivative of $f$, so it is concave on $[1, \infty)$. Why can't you just skip everything else and directly apply jensen's here to finish it off?

Because not all of $a,b,c$ are in $[1, \infty)$. We did the substitution to make them so, and we need a justification as to why that substitution increases $f(a)+f(b)+f(c)$.

For some positive real $x<y$ staisfying $xy\geq 1$ consider smoothing $(x,y)\rightarrow (x+\epsilon ,y-\epsilon )$ with $\epsilon \in \left[ 0;\frac{y-x}{2} \right].$ Claim 1. Value of $xy$ doesn't decrease. Proof. Well-known: $(x+\epsilon)(y-\epsilon)-xy=\epsilon (y-x-\epsilon)\geq 0$ $\Box$ Claim 2. Value of $(x^2+1)(y^2+1)$ doesn't decrease. Proof. Consider function $f(\epsilon )=((x+\epsilon)^2+1)((y-\epsilon)^2+1).$ Then $$f'(\epsilon)=2(x+y+(x+\epsilon)(y-\epsilon)^2+(x+\epsilon)^2(\epsilon-y))=$$$$=2(x+y+(x+\epsilon)(y-\epsilon)(y-x-2\epsilon))\stackrel{\text{AM-GM}}{\geq} 4\sqrt{xy}\geq 4>0\text{ } \Box$$ Now it suffices to consider smoothing $(a,b,c)\rightarrow \left( \frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3} \right),$ and the conclusion follows.

Let $(x, y, z)=(\tfrac{b+c}{2}, \tfrac{a+c}{2}, \tfrac{a+b}{2})$. Notice that by AM-GM, $\min(x, y, z) \ge 1$. Let $f(x)=\ln(x^2+1)$, and since $f$ is concave over $[1, \infty)$, Jensen's inequality yields \[ f\left(\frac{a+b+c}3\right)=f\left(\frac{x+y+z}3\right)\ge\frac{f(x)+f(y)+f(z)}3, \]so showing $f\left(\frac{a+b}2\right)\ge\frac{f(a)+f(b)}2$ is sufficient, which is equivalent to \[ (a^2+1)(b^2+1) \le \left( \left( \frac{a+b}{2} \right)^2+1 \right)^2. \]But subtracting the LHS from the RHS produces \[ \frac1{16} (a-b)^2 \left( a^2+6ab+b^2-8 \right), \]which is clearly nonnegative.

Fix $a+b+c$, we will show the maximum left hand side is when $a=b=c$. In order to do so, assume the variables are not equal (wlog, $a\neq b$). Replacing $a,b$ with $\frac{a+b}2, \frac{a+b}2$ yields a greater LHS (which can be proven by rearrangement and factoring), as desired.

This is a fairly straightforward smoothing problem; first, notice that it suffices to prove the two variable version of the problem, from where a simple Cauchy induction proves the case $n=4$ then $n=3$. On the other hand, the inequality $$(a^2+1)(b^2+1) \leq \left(\left(\frac{a+b}2\right)^2 + 1\right)^2$$expands to $$(a-b)^2(a^2+6ab+b^2-8) \geq 0,$$which is obviously true.

Define $f(x) = \ln(x^2 + 1)$ which is concave on $[1, \infty)$. We want to show that \[ f(a) + f(b) + f(c) \le 3f\left(\frac{a+b+c}{3}\right). \]Claim: If $ab \ge 1$, then \[ \sqrt{(a^2+1)(b^2+1)} \le \left( \frac{a+b}{2} \right)^2 + 1. \]Proof. This expands as \begin{align*} \sqrt{(a^2+1)(b^2+1)} &\le \frac{a^2 + 2ab + b^2 + 4}{4} \\ a^4 + b^4 - 10a^2b^2 + 4a^3b + 4ab^3 + 16ab - 8a^2 - 8b^2 &\ge 0 \\ (a - b)^2(a^2 + 6ab + b^2 - 8) &\ge 0 \end{align*}which follows as $ab \ge 1$. $\blacksquare$ WLOG let $a \le b \le c$. Then, by Jensen's \[ f(a) + f(b) + f(c) \le 2f\left(\frac{a+b}{2}\right) + f(c) \le 3\left(\frac{a+b+c}{3}\right) \]as $a + b \ge 2$.

Nice. I think I found long time ago a similar proof

We show the two variable inequality first. Observe that if $xy \ge 1$: \[ (x^2 + 1)(y^2 + 1) \le \left( \left( \frac{x + y}{2} \right)^2 + 1 \right)^2 \]becomes \[ \frac 1{16} (x-y)^2(x^2 + 6xy + y^2 - 8) \ge 0 \]which is true because $x^2 + 6xy + y^2 \ge 8xy \ge 1$. A la proof of AM-GM, note that the four variable inequality is true from this; indeed, \[ \frac{(a + b)(c + d)}{4} \ge 1 \iff ab + bc + cd + da \ge 4, \]which is true by assuming $\min(ab, bc, cd, da, ac, bd) \ge 1$. So we have \[ \sqrt[4]{(a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1)} \le \left( \frac{a + b + c + d}{4} \right)^2 + 1. \]Setting $d = \frac{a + b + c}{3}$, if we WLOG assume $a \ge c \ge b$, we must have that $a \ge c \ge 1$, and so $ad \ge 1$ and $ac \ge 1$, so we can apply the inequality; so we have \[ (a^2 + 1)(b^2 + 1)(c^2 + 1)\left( \left(\frac{a + b + c}{3} \right)^2 + 1 \right) \le \left( \left( \frac{a + b + c}{3} \right)^2 + 1 \right)^4 \]which rearranges into the result.

Good proof

WLOG, $a\leq b\leq c$. First, we take the natural log of both sides. If we set $f(x)=\ln x$, it suffices to show: \[\frac{f(a)+f(b)+f(c)}{3}\leq f\left(\frac{a+b+c}{3}\right).\]This is true by Jensen's inequality on the interval $[1,\infty)$, because $f$ is concave. Henceforth, assume WLOG $a<1$. Note that $\frac{a+b}{2}\geq \sqrt{ab}\geq 1$. Therefore consider $x=\frac{a+b}{2}, y=\frac{a+b}{2}$. If: \[f(a)+f(b)+f(c)\leq f(x)+f(y)+f(c),\]then: \[f(a)+f(b)+f(c)\leq 3f\left(\frac{a+b+c}{3}\right).\] If this is true, then: \[\sqrt{(a^2+1)(b^2+1)}\leq \left(\frac{a+b}{2}\right)^2+1,\]\[16(a^2b^2+a^2+b^2+1)\leq (a+b)^4+16+8(a+b)^2,\]\[(a+b)^4-16a^2b^2\geq 8(a-b)^2,\]\[(a-b)^2(a^2+6ab+b^2)\geq 8(a-b)^2,\]which is true by AM-GM, so we are done.

Generalization 1 Let $a_{1},a_{2},\cdots,a_{2^{p}-x}, \lambda $ be positive reals ($x+1\leq 2^p$ ve $x \geq 0$) such that $\min{(a_ia_j)}\geq 1$. Then prove that $$\sqrt[2^p-x]{\prod{a_{1}^2+\lambda}}\leq \left(\dfrac{\sum_{cyc}{a_{1}}}{2^p-\lambda}\right)^2+\lambda$$

used hints Assume $a\le b\le c.$ Then $((a+b)^2+4)^2-16(a^2+1)(b^2+1)=(a-b)^2((a-b)^2+8(ab-1))\ge0$ so if $f(x)=\log(x^2+1)$ then $f(a)+f(b)\le 2f\left(\frac{a+b}2\right),$ and $f$ is convex on $[1,\infty)$ and $\frac{a+b}2\ge\sqrt{ab}\ge 1$ so Jensen's gives \[\frac{f(a)+f(b)+f(c)}3\le \frac23f\left(\frac{a+b}2\right)+\frac13f(c)\le f\left(\frac{a+b+c}3\right),\]and taking $e$ to the power of both sides gives the desired result.

AZOT1 wrote: Genralization

Let $k,a_1$, $a_2$,.., $a_n$ be positive real numbers such that: $\min \{ a_ia_j \mid 1 \leq i<j\leq n \} ] \ge k$. Prove that $$\sqrt[n]{\prod_{i=1}^{n}(a_i^2+k)} \le \left(\frac{\sum_{i=1}^{n}a_i}{n}\right)^2 + k.$$ The $n$-term is a generalization however placing $k$ in there isn't a generalization. Substitute $a_1=\sqrt{\lambda} b_1$ so $min(b_ib_j)\geq 1$ and you get exact same output with the original inequality which doesn't generalize it

We do this by smoothing. Fix $a + b + c$. We claim if $ab,bc,ca \ge 1$, replacing any two variables with their average does not decrease $\sqrt[3] {(a^2 + 1)(b^2 + 1)(c^2 + 1)}$. It it sufficient to prove $(a^2 + 1)(b^2 + 1) \le (\frac 12(a + b) + 1)^2$, which is equivalent to $a^2b^2 + a^2 + b^2 \le \frac{1}{16}(a + b)^4 + \frac 12 (a+b)^2$ which is further equivalent to $a^2b^2 + \frac 12 (a - b)^2 \le \frac{1}{16}(a + b)^4$ which is further equivalent to $8(a - b)^2 \le a^4 + b^4 + 4a^3b + 4b^3a - 10a^2b^2 = (a - b)^2 (a^2 + 6ab + b^2)$, which is equivalent to $8 \le a^2 + 6ab + b^2$, which is true by $AM-GM$ and our condition. In addition, note that $\min (ab, bc, ca)$ is nondecreasing by this operation. To see this, without loss of generality let $a,b$ be the numbers operated on with $a < b$. Then $ab$ increases and $\min (bc, ac)$ goes from $ac$ to $c \frac 12 (a + b)$. Thus we can apply this operation infinitely often. Now assume some tuple $x,y,z$ satisfies the conditions of the problem and has $x + y + z = c, k = \sqrt[3]{\prod_{cyc} x^2 + 1} > (\frac c3)^2 + 1 $. Let the ratio $\frac{k}{(\frac c3)^2 + 1} = m > 1$. Then applying the operation an arbitrary number of times produces a new tuple $(p,q,r)$ such that the maximum distance from $\frac c3$ that one of the numbers is is less than some number $f$, and $\frac{\sqrt[3]{((p^2 + 1) (q^2 + 1)(r^2 + 1))}}{(\frac c3)^2 + 1} \ge m$. Then we can make $f$ as small as we want, so the maximum ratio $\frac{p^2 + 1}{(\frac c3)^2 + 1} = 1 + \frac{2\frac c3 f + f^2 + 1}{(\frac c3)^2 + 1}$ is $g$ which is a polynomial function of $f$, so $g$ can be as close to $1$ as we want. Then we must have $m \le \sqrt[3] {g^3} = g$, impossible since $g$ can be any number above $1$.

Please contact westskigamer@gmail.com if there is an error with my solution.

Taking the natural logarithm on both sides yields the equivalent inequality $$F(a, b, c) = \frac13 \sum_{cyc} \ln (a^2+1) \leq \ln\left( \left( \frac{a+b+c}{3} \right)^2+1 \right).$$But observe that the function $f(x)=\ln(x^2+1)$ is concave over $x \in [1, \infty)$ so the above inequality holds if $a, b, c \geq 1.$ It suffices to prove the inequality when otherwise. By our condition, we have that $ab, bc, ca \geq 1$ so WLOG we must have $a \leq 1 \leq b \leq c.$ Now the following lemma, for $ab \geq 1,$ holds: $$\ln(a^2+1)+\ln(b^2+1) \leq 2\ln\left( \left( \frac{a+b}{2} \right)^2+1 \right).$$The proof is straight up expansion and then simplifying. Therefore, it follows that $$F(a, b, c) \leq F \left( \frac{a+b}{2}, \frac{a+b}{2}, c \right) \leq \ln\left( \left( \frac{a+b+c}{3} \right)^2+1 \right),$$where the last inequality holds by the same Jensen's as above because $\frac{a+b}{2} \geq \sqrt{ab} \geq 1$ by AM-GM. QED