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Algebra Difficulty 4.7 AIME Prove it Ukraine

For positive numbers aa, bb with a+b=aba + b = ab prove inequality:
ab2+4+ba2+4≥12. \frac{a}{b^2 + 4} + \frac{b}{a^2 + 4} \ge \frac{1}{2}.

Solution

ab=a+b≥2ab⇒ab≥4ab = a + b \ge 2\sqrt{ab} \Rightarrow ab \ge 4, then
ab2+4+ba2+4≥ab2+ab+ba2+ab==ab(a+b)+ba(a+b)=a2+b2(a+b)2≥12. \begin{aligned} \frac{a}{b^2+4} + \frac{b}{a^2+4} &\ge \frac{a}{b^2+ab} + \frac{b}{a^2+ab} = \\ &= \frac{a}{b(a+b)} + \frac{b}{a(a+b)} = \frac{a^2+b^2}{(a+b)^2} \ge \frac{1}{2}. \end{aligned}

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