Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it

Example 29 Prove: For non-negative real numbers $a, b, c$ , we have
$a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum a^{2} \sqrt{a^{2}+24 b c}$

Solutions — 2

Solution 1

Prove that $a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum_{\mathrm{cyc}} a^{2} \sqrt{a^{2}+24 b c}$
$\Leftrightarrow \sum_{\mathrm{cyc}}\left(a^{3} b^{3}+24 a^{2} b^{2} c^{2}\right) \leqslant \sum_{\mathrm{cyc}} a^{2} b^{2} \sqrt{\left(a^{2}+24 b c\right)\left(b^{2}+24 a c\right)} .$

Since
$\sqrt{\left(a^{2}+24 b c\right)\left(b^{2}+24 a c\right)} \geqslant a b+24 c \sqrt{a b},$

it suffices to prove
$\sum_{c y c}(a b \sqrt{a b}-a b c) \geqslant 0$

By Muirhead's theorem, the above inequality is obviously true, hence the original inequality holds.

Solution 2

Prove $a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum_{\mathrm{cyc}} a^{2} \sqrt{a^{2}+24 b c}$
$\Leftrightarrow \sum_{\mathrm{cyc}}\left(a^{3} b^{3}+24 a^{2} b^{2} c^{2}\right) \leqslant \sum_{\mathrm{cyc}} a^{2} b^{2} \sqrt{\left(a^{2}+24 b c\right)\left(b^{2}+24 a c\right)} .$

Since
$\sqrt{\left(a^{2}+24 b c\right)\left(b^{2}+24 a c\right)} \geqslant a b+24 c \sqrt{a b}$

It suffices to prove
$\sum_{\mathrm{cyc}}(a b \sqrt{a b}-a b c) \geqslant 0$

By Muirhead's theorem, the above inequality is obviously true, hence the original inequality holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.