Question

Cho các số thực x, y, z \(\ge\) 1 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

CMR: \(\sqrt{x+y+z}\) \(\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\)

Answer 1

Akai Haruma

@Nguyễn Việt Lâm

@Trần Thanh Phương

Answer 2

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Leftrightarrow1-\frac{1}{x}+1-\frac{1}{y}+1-\frac{1}{z}=1\)

\(\Leftrightarrow1=\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}\ge\frac{\left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2}{x+y+z}\)

\(\Rightarrow x+y+z\ge\left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2\)

\(\Rightarrow\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\)