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x^2+y^2+z^2 + xy+yz + zx> =2(\sqrt{x} +\sqrt{y}+ \sqrt{z}) if xyz = 1, x,y,z>0

Source: Austrian Polish 1991 APMC

May 1, 2020
inequalitiesalgebra

Problem Statement

If x,y,zx,y, z are arbitrary positive numbers with xyz=1xyz = 1, prove the inequality x2+y2+z2+xy+yz+zx≥2(x+y+z)x^2+y^2+z^2 + xy+yz + zx \ge 2(\sqrt{x} +\sqrt{y}+ \sqrt{z}).
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