Question

Let xi ∈ R such that 0 < x1 ≤ x2 ≤ .....≤ xn,
n ≥ 2, and 1/(1+x1) + 1/(1+x2) +........+1/(1+xn) =1 then show that
√x1 + √x2 +........+ √xn ≥ (n-1) {1/√x1 + .....+ 1/√xn}

Answer 1

Answer on Question #78460 – Math – Algebra

Question

Let $x_1, \ldots, x_n \in \mathbb{R}, n \geq 2$, such that $0 < x_1 \leq x_2 \leq \ldots \leq x_n$, and $\frac{1}{1 + x_1} + \frac{1}{1 + x_2} + \ldots + \frac{1}{1 + x_n} = 1$, then show that $\sqrt{x_1} + \sqrt{x_2} + \ldots + \sqrt{x_n} \geq (n - 1)\left(\frac{1}{\sqrt{x_1}} + \ldots + \frac{1}{\sqrt{x_n}}\right)$.

Solution

0) This problem originally appeared on the Vojtěch Jarník Competition, in 2002.

1) Consider a case when $n = 2$.

$$1 = \frac{1}{1 + x_1} + \frac{1}{1 + x_2} = \frac{x_1 + x_2 + 2}{x_1 x_2 + x_1 + x_2 + 1} \Rightarrow x_1 x_2 = 2 - 1 = 1$$ $$\sqrt{x_1} + \sqrt{x_2} - \frac{1}{\sqrt{x_1}} - \frac{1}{\sqrt{x_2}} = \frac{(\sqrt{x_1} \sqrt{x_2} - 1) \sqrt{x_1} + (\sqrt{x_1} \sqrt{x_2} - 1) \sqrt{x_2}}{\sqrt{x_1} \sqrt{x_2}} = 0 \geq 0.$$

2) First, assume that $x_1 \leq 1$ and $i \geq 2$, then $\frac{1}{1 + x_i} \leq \frac{1}{1 + x_2} + \ldots + \frac{1}{1 + x_n} \leq 1 - \frac{1}{1 + x_1} = \frac{x_1}{1 + x_1} = \frac{1}{1 + \frac{1}{x_1}}$, then $x_i \geq 1$, and $x_i \geq 1 / x_1$.

Next, consider a sequence $a_i = \sqrt{x_i} + \frac{1}{\sqrt{x_i}} = \frac{x_i + 1}{\sqrt{x_i}}$. It is obvious that $0 < 1 \leq a_1 \leq a_2 \leq \ldots \leq a_n$ if $1 < x_1 \leq x_2 \leq \ldots \leq x_n$ and $0 \leq a_1 \leq 1 \leq a_2 \leq \ldots \leq a_n$ if $x_1 \leq 1 \leq \frac{1}{x_1} \leq x_2 \leq \ldots \leq x_n$:

Now, consider a sequence $b_i = \frac{1}{1 + x_i}$. It is obvious that $b_1 \geq b_2 \geq \ldots \geq b_n > 0$.

The Chebyshev's inequality gives that $\sum a_i * \sum b_i \equiv \sum \left( \sqrt{x_i} + \frac{1}{\sqrt{x_i}} \right) * 1 \geq n \sum a_i b_i \equiv n \sum \frac{1}{x_i}$. This completes the proof.

The equality holds if and only $a_{i} = a_{1}$ or $b_{i} = b_{1}$. It is easy to see that the latter means that $x_{1} = x_{i} = n - 1$. The former yields one more case: $x_{1} \leq 1 \leq \frac{1}{x_{1}} = x_{i}$. Since $1 = \frac{1}{1 + \frac{1}{x_{2}}} + \frac{(n - 1)}{1 + x_{2}} = \frac{n - 1 + x_{2}}{1 + x_{2}}$, it follows that $n - 1 = 1$.

**Answer:**

The statement can be proved using Chebyshev's inequality. The equality holds if and only if $x_{1} = x_{i} = n - 1$ ($n \geq 2$) or $x_{1} < 1 < \frac{1}{x_{1}} = x_{2}$ ($n = 2$).

Answer provided by https://www.AssignmentExpert.com