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Algebra Difficulty 6.5 National olympiad Prove it

Example 8.1.4 Let a,b,c≥0a, b, c \geq 0, prove: a3b2−bc+c2+b3c2−ca+a2+c3a2−ab+b2≥a+b+c\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}} \geq a+b+c

Solution

Prove: By applying the Cauchy-Schwarz inequality, we have
∑cyca3b2−bc+c2=∑cyca4a(b2−bc+c2)≥(a2+b2+c2)2∑cyca(b2−bc+c2), \sum_{c y c} \frac{a^{3}}{b^{2}-b c+c^{2}}=\sum_{c y c} \frac{a^{4}}{a\left(b^{2}-b c+c^{2}\right)} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{c y c} a\left(b^{2}-b c+c^{2}\right)},
thus it suffices to prove
(∑cyca2)2≥(∑cyca(b2−bc+c2))(∑cyca) \left(\sum_{c y c} a^{2}\right)^{2} \geq \left(\sum_{c y c} a\left(b^{2}-b c+c^{2}\right)\right)\left(\sum_{c y c} a\right)
or
∑cyca4+2∑cyca2b2≥(a+b+c)∑cyca2(b+c)−3abc∑cyca \sum_{c y c} a^{4}+2 \sum_{c y c} a^{2} b^{2} \geq (a+b+c) \sum_{c y c} a^{2}(b+c)-3 a b c \sum_{c y c} a
or
∑cyca4+abc∑cyca≥∑cyca3(b+c). \sum_{c y c} a^{4}+a b c \sum_{c y c} a \geq \sum_{c y c} a^{3}(b+c).
This is a fourth-degree Schur inequality. The equality holds if and only if a=b=ca=b=c or a=b,c=0a=b, c=0 and its permutations.

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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.