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\sqrt{(w^2 + x^2 + y^2 + z^2)/4} >=\sqrt[3]{wxy + wxz + wyz + xyz)/4}

Source: 1964 Hungary - Kürschák Competition p3

October 11, 2022
algebrainequalities

Problem Statement

Show that for any positive reals w,x,y,zw, x, y, z we have w2+x2+y2+z24≥wxy+wxz+wyz+xyz43\sqrt{\frac{w^2 + x^2 + y^2 + z^2}{4}}\ge \sqrt[3]{ \frac{wxy + wxz + wyz + xyz}{4}}
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