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Algebra Difficulty 4.9 AIME Prove it Ukraine

Let be given positive numbers a,b,ca, b, c. Prove that:
a2a2+b2+c2+b2b2+c2+a2+c2c2+a2+b2≤94(a+b+c) \frac{a}{2a^2 + b^2 + c^2} + \frac{b}{2b^2 + c^2 + a^2} + \frac{c}{2c^2 + a^2 + b^2} \le \frac{9}{4(a+b+c)}

Solution

Using the inequality a2+b2+c2≥ab+bc+caa^2 + b^2 + c^2 \ge ab + bc + ca we have
a(a+b)(a+c)+b(b+c)(b+a)+c(c+a)(c+b)≤94(a+b+c) \frac{a}{(a+b)(a+c)} + \frac{b}{(b+c)(b+a)} + \frac{c}{(c+a)(c+b)} \le \frac{9}{4(a+b+c)}
, is equivalent to
8(a+b+c)(ab+bc+ca)≤9(a+b)(b+c)(c+a) 8(a+b+c)(ab+bc+ca) \le 9(a+b)(b+c)(c+a)
and this equivalent to inequality
6abc≤a2b+a2c+b2a+b2c+c2a+c2b 6abc \le a^2b + a^2c + b^2a + b^2c + c^2a + c^2b
which follows immediately by the AM-GM inequality applied to the six numbers.

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