Classic 3 variable inequality

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Classic 3 variable inequality

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Source: Mathematical Minds 2024 P4

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#1 h Sep 29, 2024, 10:33 AM

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Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ . Prove that $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+a^3}{2}}\leqslant a^2+b^2+c^2.$
Proposed by Andrei Vila

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no_room_for_error

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no_room_for_error

#2 h Sep 29, 2024, 11:25 AM • 3 Y

Y by ytChen, ehuseyinyigit, MihaiT

AndreiVila wrote:

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ . Prove that $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+a^3}{2}}\leqslant a^2+b^2+c^2.$
Proposed by Andrei Vila

Using AM-GM and $a^2+b^2+c^2\geq 3$ :

$\begin {aligned} \sum \sqrt[3]{\frac{a^3+b^3}{2}} &= \sum \sqrt[3]{\frac{1}{2}(a+b)(a^2-ab+b^2)} \\ &\leq \frac{1}{3}\sum\left(1+\frac{a+b}{2}+a^2-ab+b^2\right)\\ &= \frac{6+2(a^2+b^2+c^2)-ab-bc-ca}{3}\\ &= \frac{3+5(a^2+b^2+c^2)}{6}\\ &\leq a^2+b^2+c^2 \end{aligned}$

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arqady

#3 h Sep 29, 2024, 12:15 PM

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AndreiVila wrote:

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ . Prove that $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+a^3}{2}}\leqslant a^2+b^2+c^2.$
Proposed by Andrei Vila

Also, $\sqrt[3]{\frac{a^3+b^3}{2}}\leq\frac{a^2+b^2}{a+b}$ helps.

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#4 h Sep 29, 2024, 12:30 PM

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Yes, Mildorf's Lemma. I was about to post it arqady

By well-known Mildorf's Inequality
$LHS=\sum_{cyc}{\sqrt[3]{\dfrac{a^3+b^3}{2}}}\leq \sum_{cyc}{\dfrac{a^2+b^2}{a+b}}=6-2\sum_{cyc}{\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}}$ From now on, Schur's Inequality can help. However, a more direct way would be showing
$\sum_{cyc}{\dfrac{a^2+b^2}{a+b}}\leq a^2+b^2+c^2$ which can be proven upon expanding or using $\dfrac{9}{8}(a+b)(b+c)(c+a)\geq (a+b+c)(ab+bc+ca)$ .

This post has been edited 1 time. Last edited by ehuseyinyigit, Sep 29, 2024, 12:31 PM

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#5 h Apr 13, 2025, 2:01 PM

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no_room_for_error wrote:

AndreiVila wrote:

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ . Prove that $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+a^3}{2}}\leqslant a^2+b^2+c^2.$
Proposed by Andrei Vila

Using AM-GM and $a^2+b^2+c^2\geq 3$ :

It would be QM AM ig

This post has been edited 1 time. Last edited by Rohit-2006, Apr 13, 2025, 2:02 PM

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