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Algebra Difficulty 7.3 National olympiad, round 2 Prove it

14. Let a,b,ca, b, c be positive numbers, no two of which are zero. If nn is a positive integer, then
2an−bn−cnb2−bc+c2+2bn−cn−anc2−ca+a2+2cn−an−bna2−ab+b2⩾0\frac{2 a^{n}-b^{n}-c^{n}}{b^{2}-b c+c^{2}}+\frac{2 b^{n}-c^{n}-a^{n}}{c^{2}-c a+a^{2}}+\frac{2 c^{n}-a^{n}-b^{n}}{a^{2}-a b+b^{2}} \geqslant 0

Solution

14. (2007.04.06) Simplify
∑2an−bn−cnb2−bc+c2=∑(an−bn)−(cn−an)b2−bc+c2=∑(1b2−bc+c2−1c2−ca+a2)(an−bn)=∑(a+b−c)(a−b)(an−bn)(b2−bc+c2)(c2−ca+a2)⩾0⇔∑(a+b−c)(a2−ab+b2)(a−b)(an−bn)⩾0\begin{array}{l} \sum \frac{2 a^{n}-b^{n}-c^{n}}{b^{2}-b c+c^{2}}=\sum \frac{\left(a^{n}-b^{n}\right)-\left(c^{n}-a^{n}\right)}{b^{2}-b c+c^{2}}= \\ \sum\left(\frac{1}{b^{2}-b c+c^{2}}-\frac{1}{c^{2}-c a+a^{2}}\right)\left(a^{n}-b^{n}\right)= \\ \sum \frac{(a+b-c)(a-b)\left(a^{n}-b^{n}\right)}{\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right)} \geqslant 0 \Leftrightarrow \\ \sum(a+b-c)\left(a^{2}-a b+b^{2}\right)(a-b)\left(a^{n}-b^{n}\right) \geqslant 0 \end{array}

By symmetry, assume a⩾b⩾ca \geqslant b \geqslant c, then a−c⩾b−c,an−cn⩾bn−cn,(a−b)⋅a-c \geqslant b-c, a^{n}-c^{n} \geqslant b^{n}-c^{n},(a-b) \cdot
(an−bn)⩾0,(b−c)(bn−cn)⩾0,(a−c)(an−cn)⩾0, therefore, we have ∑(a+b−c)(a2−ab+b2)(a−b)(an−bn)⩾(b+c−a)(b2−bc+c2)(b−c)(bn−cn)+(c+a−b)(c2−ca+a2)(a−c)(an−cn)⩾[(b+c−a)(b2−bc+c2)+(c+a−b)(c2−ca+a2)](b−c)(bn−cn)⩾[(b−a)(b2−bc+c2)+(a−b)(c2−ca+a2)](b−c)(bn−cn)⩾[(c2−ca+a2)−(b2−bc+c2)](a−b)(b−c)(bn−cn)=(a+b−c)(a−b)2(b−c)(bn−cn)⩾\begin{array}{l} \left(a^{n}-b^{n}\right) \geqslant 0,(b-c)\left(b^{n}-c^{n}\right) \geqslant 0,(a-c)\left(a^{n}-c^{n}\right) \geqslant 0, \text { therefore, we have } \\ \sum(a+b-c)\left(a^{2}-a b+b^{2}\right)(a-b)\left(a^{n}-b^{n}\right) \geqslant \\ (b+c-a)\left(b^{2}-b c+c^{2}\right)(b-c)\left(b^{n}-c^{n}\right)+ \\ (c+a-b)\left(c^{2}-c a+a^{2}\right)(a-c)\left(a^{n}-c^{n}\right) \geqslant \\ {\left[(b+c-a)\left(b^{2}-b c+c^{2}\right)+(c+a-b)\left(c^{2}-c a+a^{2}\right)\right](b-c)\left(b^{n}-c^{n}\right) \geqslant} \\ {\left[(b-a)\left(b^{2}-b c+c^{2}\right)+(a-b)\left(c^{2}-c a+a^{2}\right)\right](b-c)\left(b^{n}-c^{n}\right) \geqslant} \\ {\left[\left(c^{2}-c a+a^{2}\right)-\left(b^{2}-b c+c^{2}\right)\right](a-b)(b-c)\left(b^{n}-c^{n}\right)=} \\ (a+b-c)(a-b)^{2}(b-c)\left(b^{n}-c^{n}\right) \geqslant \end{array}

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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.