FE inequality from Iran

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FE inequality from Iran

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Source: Iran 2025 second round P5

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#1 h Apr 19, 2025, 9:20 AM • 1 Y

Y by sami1618

Find all functions $f:\mathbb{R}^+ \to \mathbb{R}$ such that for all $x,y,z>0$
$3(x^3+y^3+z^3)\geq f(x+y+z)\cdot f(xy+yz+xz) \geq (x+y+z)(xy+yz+xz).$

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#2 h Apr 19, 2025, 10:15 AM • 1 Y

Y by sami1618

Answers are $f(x)=x$ and $f(x)=-x$ which clearly hold. Let $f(x)=xg(x)$ . We have
$\frac{3(x^3+y^3+z^3)}{(x+y+z)(xy+yz+zx)}\geq g(x+y+z)g(xy+yz+zx)\geq 1$ Lemma: For positive reals $a,b$ if $a^2\geq 3b$ , then there exists positive reals $x+y+z=a$ and $xy+yz+zx=b$ .
Proof: Omitted.

Pick $x=y=z$ to get $g(3x)g(3x^2)=1$ . If $3a^2\geq 3a\geq 3b\geq 3$ or $3a^2\geq 3a\geq 3\geq 3b$ , then by the lemma $g(3a)g(3b)\geq 1$ and $g(3a^2)g(3b)\geq 1$ and multiplying these give $g(3b)^2\geq 1$ thus, $|g(x)|\geq 1$ for all positive reals. Since $g(3x)g(3x^2)=1$ , we get $|g(3x)|=1$ thus, $|g(x)|=1$ . If $g(3a)=-1$ for some $a>1$ , then $g(3a^{2^k})=-1$ . Since $g(3a^{2^k})g(b)\geq 1$ for $9a^{2^{k+1}}\geq 3b$ which holds for sufficiently large $k$ , we see that $g(b)\leq -1$ and since $|g(b)|=1$ , $g\equiv -1$ holds.
If $g(3a)=1$ for all $a>1$ , then $g(3a^{2^k})=1$ . Since $g(3a^{2^k})g(b)\geq 1$ which holds for sufficiently large $k$ , we observe $g(b)\geq 1$ thus, $g\equiv 1$ as desired. $\blacksquare$

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#4 h Apr 20, 2025, 2:15 PM

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Answer: $f(x)=x$ or $f(x)=-x$ .

Solution. It is straightforward to check that the above functions satisfy the inequality using Muirhead. Let $f(x)$ satisfy the inequality. By letting $(x,y,z)=(1,1,1)$ we get that $f(3)=\pm 3$ . Notice that the function $-f$ also satisfies the inequality, so we can suppose without loss of generality that $f(3)=3$ .

Claim 1. $f(x)=x$ for all $x\in (0,3)$ .
Proof. Let $t\in(0,1)$ . Setting $(x,y,z)=(1-t,1-t,1+2t)$ gives $f(3-3t^2)\geq 3-3t^2$ so $f(x)\geq x$ for all $x\in (0,3)$ . Letting $(x,y,z)=(s,s,s)$ for some $s\in (0,1)$ gives $9s^3\geq f(3s)f(3s^2)$ . But since $f(3s)\geq 3s$ and $f(3s^2)\geq 3s^2$ , it follows that $f(3s)=3s$ , proving the claim.

Claim 2. $f(x)=x$ for all $x\in (3,\infty)$ .
Proof. Any number $x\in(3,\infty)$ can be expressed as $t+2t^{-1}$ for some $t\in(2,\infty)$ . Setting $(x,y,z)=(t,t^{-1},t^{-1})$ gives that $f(x)f(2+t^{-2})\geq x(2+t^{-2})$ . Since $2+t^{-2}\in(0,3)$ , it follows that $f(x)\geq x$ for all $x\in (3,\infty)$ . Letting $(x,y,z)=(s,s,s)$ for some $s\in (1,\infty)$ gives that $9s^3\geq f(3s)f(3s^2)$ . But since $f(3s)\geq 3s$ and $f(3s^2)\geq 3s^2$ , it follows that $f(3s)=3s$ , proving the claim.

Combining both Claims gives that $f(x)=x$ for all $x\in \mathbb{R}^+$ , as required.

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#5 h Apr 22, 2025, 1:15 PM • 1 Y

Y by blackidea

we show the inequality of the problem with $p(x,y,z)$ :
$p(x,x,x) : 9x^3 \geq f(3x)f(3x^2) \geq 9x^3 \implies f(3x)f(3x^2) = 9x^3$ Easily we get that:
$f(a)f\left(\frac{a^2}{3}\right) = \frac{a^3}{3} \quad (\forall a\in\mathbb R_{> 0})$ Define $h(x) := \left|\frac{f(x)}{x}\right|$ . Then we have
$h(a)h\left(\frac{a^2}{3}\right) = 1 \quad (\forall a\in\mathbb R_{> 0})$ On the other hand we have
$p\left(x,\frac{1}{x},1\right) : \left(f\left(x + \frac{1}{x} + 1\right)\right)^2 \geq \left(x + \frac{1}{x} + 1\right)^2$ and because of we have $x + \frac{1}{x} + 1 \geq 3$ ,
$|f(a)| \geq a \quad (\forall a \geq 3) \implies h(a) \geq 1 \quad (\forall a \geq 3)$ If $a \geq 3$ then $\frac{a^2}{3} \geq 3$ which means that $h\left(\frac{a^2}{3}\right) \geq 1$ .
If we multiply this inequality with $h(a) \geq 1$ we get that
$h(a)h\left(\frac{a^2}{3}\right) \geq 1 \quad (\forall a \geq 3)$ But we proved that
$h(a)h\left(\frac{a^2}{3}\right) = 1 \quad (\forall a \in \mathbb{R}_{>0})$ Therefore, for all $a \geq 3$ we have $h(a) = 1$ .
Now choose $x, y, z$ such that $\sum x = 3$ . Then $\sum xy \leq 3$ because of
$\left(\sum x\right)^2 \geq 3\sum xy$ And actually we can prove that for all $t \in (0, 3]$ we can choose $x, y, z$ such that $\sum x = 3$ and $\sum xy = t$ . For proving that we only have to solve the system below:
$\begin{cases} x + y + z = 3, \\ xy + yz + zx = t \quad (0 < t \leq 3) \end{cases}$ If we put those $x, y, z$ into the question's inequality we get that:
$f\left(\sum x\right)f\left(\sum xy\right) \geq \left(\sum x\right)\left(\sum xy\right) \implies \left|f\left(\sum x\right)\right|\left|f\left(\sum xy\right)\right| \geq \left|\sum x\right|\left|\sum xy\right|$ $\implies \left|\frac{f\left(\sum x\right)}{\sum x}\right|\left|\frac{f\left(\sum xy\right)}{\sum xy}\right| \geq 1$ And because $\sum x = 3$ we have that:
$h(a) \geq 1 \quad (\forall a \leq 3)$ Now if $a \leq 3$ then $\frac{a^2}{3} \leq 3$ , so
$h(a)h\left(\frac{a^2}{3}\right) \geq 1 \implies h(a) = 1 \quad (\forall a \leq 3)$ so for all positive real numbers $a$ , $h(a) = 1$ , which means that
$\left|\frac{f(x)}{x}\right| = 1 \implies |f(x)| = x \quad (\forall x \in \mathbb{R}_{>0})$ $\implies f(x) = \begin{cases} x & \text{if } x \in A \\ -x & \text{if } x \in B \end{cases}$ And $A, B$ are a partition of positive real numbers.
Let $1 \in X$ such that $X$ is $A$ or it is $B$ . Then all of the numbers $\frac{1}{3} \geq$ are in $X$ too.
Because if one of the numbers $\frac{1}{3} \geq$ does not belong to $X$ (let that number be $s$ ), then we can find $x, y, z$ such that $\sum x = 1$ and $\sum xy = s$ . Now if we put these $x, y, z$ in the main inequality we get that:
$0 \geq f\left(\sum x\right)f\left(\sum xy\right) \geq \left(\sum x\right)\left(\sum xy\right)$ which is a contradiction.
So all of the numbers $\frac{1}{3} \geq$ and one are in $X$ .
Similarly we can show that $2$ and all the numbers $\frac{4}{3} \geq$ are in the same set, but we know that $1$ and all the numbers $\frac{1}{3} \geq$ are in $X$ , so $2$ and all the numbers $\frac{4}{3} \geq$ are in $X$ .
And if we do it for all natural numbers we get that all the real numbers are in $X$ . So $A$ or $B = \emptyset$ (but not at the same time).
$\text{Therefore, the answers are: } \begin{cases} f(x) = x & \forall x > 0 \\ f(x) = -x & \forall x > 0 \end{cases}$ Now we only have to check this inequality:
$3\sum x^3 \geq f\left(\sum x\right)f\left(\sum xy\right)$ This is true because:
$\begin{align*} x^3 + y^3 &= (x+y)(x^2 - xy + y^2) \\ &\geq xy(x+y) \\ \implies 2\sum x^3 &\geq \sum x^2y + \sum xy^2 \\ \sum x^3 &\geq 3xyz \end{align*}$ Therefore:
$\begin{align*} 3\sum x^3 &\geq \left(\sum x\right)\left(\sum xy\right) \\ &= f\left(\sum x\right)f\left(\sum xy\right) \end{align*}$ and we are done

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