Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

Example $1.8 a, b, c \geqslant 0$ , prove that
$\sqrt{2 a^{2}+5 a b+2 b^{2}}+\sqrt{2 a^{2}+5 a c+2 c^{2}}+\sqrt{2 b^{2}+5 b c+2 c^{2}} \leqslant 3(a+b+c)$

Solution

Prove that by AM-GM inequality,
$2 a^{2}+5 a b+2 b^{2}=(2 a+b)(2 b+a) \leqslant \frac{1}{4}(2 a+b+2 b+a)^{2}=\frac{9}{4}(a+b)^{2}$

Therefore,
$\sum \sqrt{2 a^{2}+5 a b+2 b^{2}} \leqslant \frac{3}{2} \sum(a+b)=3(a+b+c)$

Equality holds if and only if $a=b=c$ .

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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.