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(px + qy)^2 <= px^2 + qy^2 if 0 < p, 0 < q, p +q < 1

Source: Spanish Mathematical Olympiad 1971 P3

December 5, 2022
algebrainequalities

Problem Statement

If 0<p0 < p, 0<q0 < q and p+q<1p +q < 1 prove (px+qy)2px2+qy2(px + qy)^2 \le px^2 + qy^2