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Problem 1210

National olympiad, first round

Algebra Difficulty 6.3 Prove it

46. Let non-negative real numbers $a, b, c$ satisfy $a^{2} \leqslant b^{2}+c^{2}, b^{2} \leqslant c^{2}+a^{2}, c^{2} \leqslant a^{2}+b^{2}$ , prove: $(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)\left(a^{3}+b^{3}+c^{3}\right) \geqslant 4\left(a^{6}+b^{6}+c^{6}\right)$ . (11th Japanese Mathematics

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

46. By Cauchy-Schwarz inequality,
$(a+b+c)\left(a^{3}+b^{3}+c^{3}\right) \geqslant\left(a^{2}+b^{2}+c^{2}\right)^{2}$

It suffices to prove
$\left(a^{2}+b^{2}+c^{2}\right)^{3} \geqslant 4\left(a^{6}+b^{6}+c^{6}\right)$

By the identity $(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(z+x)$ , it suffices to prove
$\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right) \geqslant a^{6}+b^{6}+c^{6}$

That is, to prove
$2 a^{2} b^{2} c^{2}+a^{2}\left(b^{2}+c^{2}\right)+b^{2}\left(c^{2}+a^{2}\right)+c^{2}\left(a^{2}+b^{2}\right) \geqslant a^{6}+b^{6}+c^{6}$

By the given conditions, the inequality clearly holds.

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