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Algebra Difficulty 5.0 AIME Prove it Mongolia

For any positive real numbers aa, bb, cc, dd, prove that
(a+b)2+(c+d)2(a2+c2)(b+d)+(a+c)(b2+d2)≤1a+c+1b+d. \frac{(a+b)^2 + (c+d)^2}{(a^2+c^2)(b+d) + (a+c)(b^2+d^2)} \le \frac{1}{a+c} + \frac{1}{b+d}.

Solution

Clearly it suffices to prove
(a2+c2a+c+b2+d2b+d)(a+c+b+d)≥(a+b)2+(c+d)2. \left( \frac{a^2 + c^2}{a+c} + \frac{b^2 + d^2}{b+d} \right) (a+c+b+d) \ge (a+b)^2 + (c+d)^2.
By the Cauchy-Schwartz and Minkowski inequalities, we have
(a2+c2a+c+b2+d2b+d)(a+c+b+d)≥(a2+c2+b2+d2)2≥(a+b)2+(c+d)2. \left( \frac{a^2 + c^2}{a+c} + \frac{b^2 + d^2}{b+d} \right) (a+c+b+d) \geq \left( \sqrt{a^2+c^2} + \sqrt{b^2+d^2} \right)^2 \geq (a+b)^2 + (c+d)^2.
It is easy to check that equality holds if and only if ad=bcad = bc.

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