<s>[INST] Prove that $\sqrt{x+y+z}\ge{\sqrt{x-1}+\sqrt{z-1}}$. Let $x,y,z>1$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that 
$$\sqrt{x+y+z}\ge{\sqrt{x-1}+\sqrt{z-1}}$$.
I took $x=\sec^2{a}$, $y=\sec^2{b}$, $z=\sec^2{c}$ but it was not useful. Then I took $x-1=a$ and similarly $b$ and $c$. On simplifying I got
$$b+1\ge{2\sqrt{ac}}$$
 [/INST] Applying Cauchy the following is true. $$(x+y+z)(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})\ge (\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$$
However, $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1$ from the condition in the problem. 
$\therefore$ $\sqrt{x+y+z}\ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$. 
However $\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\ge \sqrt{x-1}+\sqrt{z-1}$. Our proof is done. 
</s>