I have studied the following theorems on inequality: the QAGH (quadratic-arithmetic-geometric-harmonic) mean inequality, the triangle inequality, Cauchy Schwarz inequality, Weierstrass inequality and Chebycheff's sum inequality which states that If $a_1,a_2,\ldots a_n$ are real numbers such that (i) $$a_1\le a_2\le\ldots \le a_n$$ and $$b_1\le b_2 \le \ldots \le b_n$$, then: $$n(a_1b_1 + a_2b_2 +\ldots + a_nb_n) \ge (a_1 + a_2 + \cdots + a_n)$$
(ii) $$a_1\ge a_2\ge\ldots\ge a_n$$ and $$b_1\le b_2 \le \ldots \le b_n$$, then: $$n(a_1b_1 + a_2b_2 +\ldots + a_nb_n) \le (a_1 + a_2 + \cdots + a_n)$$
One question from my assignment for which I need help is: If $$0\le x_1\le x_2\le\ldots\le x_n,$$ and it is given that $$\frac{1}{1+x_1} +\frac{1}{1+x_2}+\ldots+\frac{1}{1+x_n} = 1$$ Prove that: $$ (\sqrt{x_1} + \sqrt{x_2} +\cdots + \sqrt{x_n}) \ge (n - 1) \left(\frac{1}{\sqrt {x_1} } +\frac{1}{\sqrt{x_2} } + \cdots + \frac{1}{\sqrt{x_n} }\right) $$ Where n is a natural number, $ n\ge 2$ I have tried in many ways to prove this statement using the theorems I know, without any success. Could you please help me?
