MathDB

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Part of 2019 Grand Duchy of Lithuania

Problems(1)

sum \sqrt{z + xy} >=\sqrt{xyz}+\sqrt{x }+\sqrt{y} +\sqrt{z} if 1/x+1/y+1/z=1

Source: 2019 Grand Duchy of Lithuania, Mathematical Contest p1 (Baltic Way TST)

10/3/2020
Let x,y,zx, y, z be positive numbers such that 1x+1y+1z=1\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1. Prove that x+yz+y+zx+z+xy≥xyz+x+y+z\sqrt{x + yz} +\sqrt{y + zx} +\sqrt{z + xy} \ge\sqrt{xyz}+\sqrt{x }+\sqrt{y} +\sqrt{z}
algebrainequalities