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Algebra Difficulty 6.2 National olympiad Prove it

25. Let a,b,ca, b, c be positive numbers. Prove that
1a2+2bc+1b2+2ca+1c2+2ab>2ab+bc+ca\frac{1}{a^{2}+2 b c}+\frac{1}{b^{2}+2 c a}+\frac{1}{c^{2}+2 a b}>\frac{2}{a b+b c+c a}

Solution

25. (2007.04.06) To prove the original inequality, it suffices to show that the left-hand side minus the right-hand side after clearing the denominators is not less than zero. Let s1=∑a,s2=∑bc,s3=abcs_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c, then
∑bc⋅∑(b2+2ca)(c2+2ab)−2∏(a2+bc)=(2s12s22−3s23)−2(4s13s3−18s1s2s3+2s23+27s32)=2(s12s22−4s23+18s1s2s3−4s13s3−27s32)+s23⩾0\begin{array}{l} \sum b c \cdot \sum\left(b^{2}+2 c a\right)\left(c^{2}+2 a b\right)-2 \prod\left(a^{2}+b c\right)= \\ \left(2 s_{1}^{2} s_{2}^{2}-3 s_{2}^{3}\right)-2\left(4 s_{1}^{3} s_{3}-18 s_{1} s_{2} s_{3}+2 s_{2}^{3}+27 s_{3}^{2}\right)= \\ 2\left(s_{1}^{2} s_{2}^{2}-4 s_{2}^{3}+18 s_{1} s_{2} s_{3}-4 s_{1}^{3} s_{3}-27 s_{3}^{2}\right)+s_{2}^{3} \geqslant \\ 0 \end{array}
(s12s22−4s23+18s1s2s3−4s13s3−27s32=(a−b)2(b−c)2(c−a)2⩾0\left(s_{1}^{2} s_{2}^{2}-4 s_{2}^{3}+18 s_{1} s_{2} s_{3}-4 s_{1}^{3} s_{3}-27 s_{3}^{2}=(a-b)^{2}(b-c)^{2}(c-a)^{2} \geqslant 0\right., see the solution to Problem 40 in Chapter 8 for reference.

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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.