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Algebra Difficulty 6.2 National olympiad Prove it

Example 1.3.5. Let a,b,ca, b, c be the side lengths of a triangle. Prove that
a2+b2a2+c2+c2+a2c2+b2+b2+c2b2+a2≥a+ba+c+a+cb+c+b+cb+a. \frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\frac{c^{2}+a^{2}}{c^{2}+b^{2}}+\frac{b^{2}+c^{2}}{b^{2}+a^{2}} \geq \frac{a+b}{a+c}+\frac{a+c}{b+c}+\frac{b+c}{b+a} .

Solution

SOLUTION. It is easy to rewrite the inequality in the following form
(a−b)2M+(c−a)(c−b)N≥0(a-b)^{2} M+(c-a)(c-b) N \geq 0
where
M=(a+b)2(a2+c2)(b2+c2)−1(a+c)(b+c)N=(a+c)(b+c)(a2+b2)(a2+c2)−1(a+c)(a+b)\begin{array}{l} M=\frac{(a+b)^{2}}{\left(a^{2}+c^{2}\right)\left(b^{2}+c^{2}\right)}-\frac{1}{(a+c)(b+c)} \\ N=\frac{(a+c)(b+c)}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}-\frac{1}{(a+c)(a+b)} \end{array}

WLOG, assume that c=min⁡{a,b,c}c=\min \{a, b, c\}. Clearly, M≥0M \geq 0 and N≥0N \geq 0 since
(a+c)2(b+c)(a+b)−(a2+b2)(a2+c2)=a3(b+c−a)+3a2bc+3abc2+a2c2+c3(a+b)≥0(a+c)^{2}(b+c)(a+b)-\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)=a^{3}(b+c-a)+3 a^{2} b c+3 a b c^{2}+a^{2} c^{2}+c^{3}(a+b) \geq 0

We are done. Equality holds for a=b=ca=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.