Question 45
Let $a, b, c$ be positive real numbers such that
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2
$$
Prove that
$$
\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2} \geqslant \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}
$$
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2
$$
Prove that
$$
\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2} \geqslant \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}
$$