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Algebra Difficulty 4.8 AIME Prove it Ukraine
Numbers a, b, c, d satisfy: ab+cd>0, ac+bd>0, and a2+d2=c2+b2. Prove that ad+bc>0.
Solution
(ab+cd)(ac+bd)=a2bc+ac2d+ab2d+bcd2=bc(a2+d2)+ad(c2+b2)=bc(a2+d2)+ad(a2+d2)=(a2+d2)(bc+ad)>0,
which implies the required inequality.
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