Inequality. ${{\sqrt{a}+\sqrt{b}+\sqrt{c}} \over {2}} \ge {{1} \over {\sqrt{a}}} + {{1} \over {\sqrt{b}}} + {{1} \over {\sqrt{c}}}$ Question. If ${{a} \over {1+a}}+{{b} \over {1+b}}+{{c} \over {1+c}}=2$ and $a$, $b$, $c$ are all positive real numbers, prove that
$${{\sqrt{a}+\sqrt{b}+\sqrt{c}} \over {2}} \ge {{1} \over {\sqrt{a}}} + {{1} \over {\sqrt{b}}} + {{1} \over {\sqrt{c}}}$$
My approach. If we let
$$x:=\frac{1}{1+a}, y:=\frac{1}{1+b}, z:=\frac{1}{1+c}$$
Then we can know that
$${{a} \over {1+a}}+{{b} \over {1+b}}+{{c} \over {1+c}}=2 \Leftrightarrow abc=a+b+c+2 \Leftrightarrow x+y+z=1$$ (By definition of x, y, z)
And,
$$a=\frac{1}{x}-1=\frac{1-x}{x}=\frac{y+z}{x} (\because x+y+z=1)$$
$$\therefore (a, b, c)=(\frac{y+z}{x}, \frac{z+x}{y}, \frac{x+y}{z})$$
T.S. $$\sum_{cyc}\frac{\sqrt{a}}{2}>\sum_{cyc}\frac{1}{\sqrt{a}}$$
$$ \Leftrightarrow \sum_{cyc}(\sqrt{a}-\frac{2}{\sqrt{a}}) \ge 0$$
$$ \Leftrightarrow \sum_{cyc}(\sqrt{\frac{y+z}{x}}-2\sqrt{\frac{x}{y+z}}) \ge 0$$
$$ \Leftrightarrow \sum_{cyc}\frac{(y-x)+(z-x)}{\sqrt{x(y+z)}}\ge 0$$
$$ \Leftrightarrow \sum_{cyc}(x-y)(\frac{1}{\sqrt{y(z+x)}}-\frac{1}{\sqrt{x(y+z)}})\ge 0$$
$$ \Leftrightarrow \sum_{cyc}(x-y)\frac{\sqrt{x(y+z)}-\sqrt{y(z+x)}}{\sqrt{xy(x+z)(y+z)}} \ge 0$$
But I don't know the next stage.
What should I do?

Now, use $$\sqrt{x(y+z)}-\sqrt{y(x+z)}=\frac{x(y+z)-y(x+z)}{\sqrt{x(y+z)}+\sqrt{y(x+z)}}=\frac{z(x-y)}{\sqrt{x(y+z)}+\sqrt{y(x+z)}}.$$
