A tricky Inequality problem $ \dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1;\ a_1 , a_2 , \ldots , a_n > 0 
$  show that  $  \sqrt{a_1} + \sqrt{a_2} + \cdots + \sqrt{a_n} \ge (n-1) \left(\dfrac{1}{\sqrt{a_1}}+ \cdots + \dfrac{1}{\sqrt{a_n}}\right)$
I have tried AM - GM and this problem is from Inequalities a mathematical Olympiad approach.
 A: Here's a method that only involves algebraic manipulations and does not require the AM-GM inequality or any other more advanced inequalities. Not the most elegant of solutions however, but I believe it works. 
$\dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1 \Rightarrow \left(1-\frac{a_1}{1+a_1}\right)+\left(1-\frac{a_2}{1+a_2}\right)+...+\left(1-\frac{a_n}{1+a_n}\right)=1$
Hence, we have: $\dfrac{a_1}{1+a_1}+\dfrac{a_2}{1+a_2}+...+\dfrac{a_n}{1+a_n}=n-1$. 
Denote the inequality that is to be proven by $(1)$.
Now, $$(1) \iff \sum_{i=1}^{n}\sqrt{a_i}-(n-1)\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}\ge0 $$
We use a small trick and write the above inequality as:
$$(1) \iff \sum_{j=1}^{n}\frac{1}{1+a_j}  \sum_{i=1}^{n}\sqrt{a_i}-\sum_{j=1}^{n}\frac{a_j}{1+a_j}\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}\ge0$$
$$ \iff \sum_{j=1}^{n}\sum_{i=1}^{n}\left(\frac{1}{1+a_j}\sqrt{a_i}-\frac{a_j}{1+a_j}\frac{1}{\sqrt{a_i}}\right)\ge 0$$
$$ \iff \sum_{j=1}^{n}\sum_{i=1}^{n}\frac{a_i-a_j}{(1+a_j)\sqrt{a_i}} \ge 0 \hspace{70pt}$$
Note that, when $i=j$, $a_i-a_j=0$. Hence:
$$(1) \iff \sum_{i>j} \left( \frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}+\frac{a_j-a_i}{(1+a_i)\sqrt{a_j}}   \right) \ge 0$$
$$\hspace{70pt} \iff \sum_{i>j} \frac{(a_i-a_j)(1+a_i)\sqrt{a_j}+(a_j-a_i)(1+a_j)\sqrt{a_i}}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$ \iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right) \left(\sqrt{a_i}+\sqrt{a_j}\right) (1+a_i) \sqrt{a_j} -  \left(\sqrt{a_i}-\sqrt{a_j}\right) \left(\sqrt{a_i}+\sqrt{a_j}\right) (1+a_j) \sqrt{a_i}}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$\iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right)\left( \sqrt{a_i}+\sqrt{a_j}  \right)  \left[(1+a_i)\sqrt{a_j}-\sqrt{a_i}(1+a_j)\right]}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$\iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right)\left( \sqrt{a_i}+\sqrt{a_j}  \right)  \left[ - \left(\sqrt{a_i}-\sqrt{a_j}\right) + \sqrt{a_i}\sqrt{a_j}\left(\sqrt{a_i}-\sqrt{a_j}\right)       \right]}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$\iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right)^2\left(\sqrt{a_i}+\sqrt{a_j}\right)\left(\sqrt{a_i}\sqrt{a_j}-1\right)}{(1+a_i)\left(1+a_j\right)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
Indeed, it suffices for us to prove that $\sqrt{a_i}\sqrt{a_j} \ge 1$. We argue as follows:
$$1 \ge \frac{1}{1+a_i}+\frac{1}{1+a_j} \Rightarrow 1 \ge \frac{2+a_i+a_j}{1+a_i+a_j+a_ia_j} \Rightarrow 1+a_i+a_j+a_ia_j \ge 2+a_i+a_j$$
$$\Rightarrow a_ia_j \ge 1 \Rightarrow \sqrt{a_i}\sqrt{a_j} \ge 1, \hspace{50pt}$$ and we are done, since all the terms in our sum are shown to be non-negative.
