Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$

This is Exercice 1.48 in the book "Inequalities-A mathematical Olympiad approach".

Attempt

I tried using HM-GM and I got $\left ( \dfrac{1}{x_1x_2\cdots x_n}\right)^{\frac{1}{2n}} \geq \dfrac{n}{\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}}} \implies \dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_1}}+\cdots+\dfrac{1}{\sqrt{x_n}} \geq n(x_1x_2 \cdots x_n)^{\frac{1}{2n}}$. But I get stuck here and don't know if this even helps.

Solution 1:

Note that $\displaystyle \sum_{i = 1}^n \dfrac{1}{1+a_i} = 1 \implies \sum_{i = 1}^n \dfrac{a_i}{1+a_i} = n-1$. Then, $$\displaystyle \sum_{i = 1}^n \sqrt{a_i}-(n-1)\sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \sum_{i = 1}^n \dfrac{1}{1+a_i}\sum \sqrt{a_i} - \sum_{i = 1}^n \dfrac{a_i}{1+a_i} \sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \displaystyle \sum_{i,j} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} = \sum_{i>j} \dfrac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}$$

But $1 \geq \dfrac{1}{1+a_i}+\dfrac{1}{1+a_j} = \dfrac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$ thus $a_ia_j \geq 1$. Hence the terms of the last sum are positive.

Solution 2:

Rewrite our inequality in the following form: $\sum\limits_{i=1}^n\frac{x_i+1}{\sqrt{x_i}}\sum\limits_{i=1}^n\frac{1}{x_i+1}\geq n\sum\limits_{i=1}^n\frac{1}{\sqrt{x_i}}$, which is Chebyshov.