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Algebra Difficulty 4.8 AIME Prove it Ukraine

Numbers aa, bb, cc, dd satisfy: ab+cd>0ab + cd > 0, ac+bd>0ac + bd > 0, and a2+d2=c2+b2a^2 + d^2 = c^2 + b^2. Prove that ad+bc>0ad + 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,(ab + cd)(ac + bd) = a^2bc + ac^2d + ab^2d + bcd^2 = bc(a^2 + d^2) + ad(c^2 + b^2) = bc(a^2 + d^2) + ad(a^2 + d^2) = (a^2 + d^2)(bc + ad) > 0,
which implies the required inequality.

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