Inequality problem: Application of Cauchy-Schwarz inequality Let $a,b,c \in (1, \infty)$ such that $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=2$. Prove that:
$$
\sqrt {a-1} + \sqrt {b-1} + \sqrt {c-1} \leq \sqrt {a+b+c}.
$$
This is supposed to be solved using the Cauchy inequality; that is, the scalar product inequality.

Let $a=x+1$, $b=y+1$ and $c=z+1$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{x+y+z+3}$$ or
$\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\leq\frac{3}{2}$.
But in another hand, the condition geves
$$\sum_{cyc}\frac{1}{x+1}=2$$ or
$$\sum_{cyc}\left(\frac{1}{x+1}-1\right)=-1$$ or
$$\sum_{cyc}\frac{x}{x+1}=1.$$
Thus, by C-S
$$1=\sum_{cyc}\frac{x}{x+1}\geq\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{x+y+z+3},$$
which gives $\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\leq\frac{3}{2}$.
Done!
