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Problem 1310

National olympiad, first round

Algebra Difficulty 6.6 Prove it

Problem 33. Let $a, b, c$ be non-negative real numbers and $a+b+c=3$ . Prove that
$a \sqrt{1+b^{3}}+b \sqrt{1+c^{3}}+c \sqrt{1+a^{3}} \leq 5$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. By AM-GM inequality, we deduce that
$\sum_{c y c} a \sqrt{1+b^{3}}=\sum_{c y c} a \sqrt{(1+b)\left(1-b+b^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(1+b^{2}\right)$

It remains to prove that
$a b^{2}+b c^{2}+c a^{2} \leq 4$

WLOG, we may suppose that $b$ is the middle number between $a, b, c$ . That means $a(b-a)(b-c) \leq 0$ , or $a b^{2}+a^{2} c \leq a b c+a^{2} b$ . It then suffices to prove that
$a b c+a^{2} b+b c^{2} \leq 4 \Leftrightarrow b\left(a^{2}+a c+c^{2}\right) \leq 4$

According to AM-GM inequality, we have
$b\left(a^{2}+a c+c^{2}\right) \leq b(a+c)^{2}=4 b \cdot \frac{(a+c)}{2} \cdot \frac{(a+c)}{2} \leq 4\left(\frac{a+b+c}{3}\right)^{3}=4$

We are done. Equality holds for $a=1, b=2, c=0$ and its permutations.

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