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Algebra Difficulty 6.1 National olympiad Prove it

36. Let a,b,ca, b, c be non-negative real numbers, prove that
b+ca2+bc+c+ab2+ca+a+bc2+ab≥4\frac{b+c}{\sqrt{a^{2}+b c}}+\frac{c+a}{\sqrt{b^{2}+c a}}+\frac{a+b}{\sqrt{c^{2}+a b}} \geq 4 (Pham Kim Hung)

Solution

Prove: By applying the Hölder's inequality, we have
(∑cycb+ca2+bc)2(∑cyc(b+c)(a2+bc))≥8(∑cyca)3\left(\sum_{c y c} \frac{b+c}{\sqrt{a^{2}+b c}}\right)^{2}\left(\sum_{c y c}(b+c)\left(a^{2}+b c\right)\right) \geq 8\left(\sum_{c y c} a\right)^{3}

Therefore, it suffices to prove
(a+b+c)3≥4∑cyca2(b+c)⇔6abc+∑cyca3≥∑cyca2(b+c)(a+b+c)^{3} \geq 4 \sum_{c y c} a^{2}(b+c) \Leftrightarrow 6 a b c+\sum_{c y c} a^{3} \geq \sum_{c y c} a^{2}(b+c)

This is true, as it can be derived from the third-degree Schur's inequality:
3abc+∑cyca3≥∑cyca2(b+c)3 a b c+\sum_{c y c} a^{3} \geq \sum_{c y c} a^{2}(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.