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ASU 543 All Soviet Union MO 1991 (x+y+z)^2/3>=x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}

Source:

August 14, 2019
algebrainequalities

Problem Statement

Show that (x+y+z)23xyz+yzx+zxy\frac{(x + y + z)^2}{3} \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy} for all non-negative reals x,y,zx, y, z.