Question

Let xi belongs to R such that 0<x1≤x2≤....≤xn, n≥2 and,
1/(1+x1) + 1/(1+x2) + ....+1/(1+xn)=n
Then show that √x1+√x2+...+√xn ≥ (n-1)(1/√x1+.....+1/√xn)
Solve using laws and theorem of inequality

Answer 1

Answer on Question #79989 – Math – Algebra

Question

Let $x$ belongs to $R$ such that $0 \leq x_1 \leq x_2 \leq \ldots \leq x_n, n \geq 2$ and $1/(1 + x_1) + 1/(1 + x_2) + \ldots + 1/(1 + x_n) = 1$, then show that

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

Solve using inequalities.

Solution

Let $\frac{1}{1 + x_i} = a_i$, for $i = 1,2,\ldots,n$. Then $\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} \frac{1}{1 + x_i} = 1$.

We have that

$$\sqrt{x_1} + \sqrt{x_2} + \cdots + \sqrt{x_n} \geq (n - 1)\left(1 / \sqrt{x_1} + \cdots + 1 / \sqrt{x_n}\right)$$ $$\begin{array}{l} \sqrt{x_i} = \sqrt{\frac{1}{a_i} - 1} = \sqrt{\frac{1 - a_i}{a_i}} \\ \sqrt{\frac{1 - a_i}{a_i}} + \sqrt{\frac{a_i}{1 - a_i}} = \sqrt{\frac{(1 - a_i)^2}{a_i(1 - a_i)}} + \sqrt{\frac{a_i^2}{a_i(1 - a_i)}} = \sqrt{\frac{1}{a_i(1 - a_i)}} \end{array}$$ $$\begin{array}{l} \sum_{i=1}^{n} \sqrt{\frac{1 - a_i}{a_i}} \geq (n - 1)\sum_{i=1}^{n} \sqrt{\frac{a_i}{1 - a_i}} \Longleftrightarrow \\ \Longleftrightarrow \sum_{i=1}^{n} \left(\sqrt{\frac{1 - a_i}{a_i}} + \sqrt{\frac{a_i}{1 - a_i}}\right) \geq n \sum_{i=1}^{n} \sqrt{\frac{a_i}{1 - a_i}} \Longleftrightarrow \\ \Longleftrightarrow \sum_{i=1}^{n} \sqrt{\frac{1}{a_i(1 - a_i)}} \geq n \sum_{i=1}^{n} \sqrt{\frac{a_i}{1 - a_i}} \Longleftrightarrow \\ \Longleftrightarrow \left(\sum_{i=1}^{n} a_i\right) \left(\sum_{i=1}^{n} \sqrt{\frac{1}{a_i(1 - a_i)}}\right) \geq n \sum_{i=1}^{n} \sqrt{\frac{a_i}{1 - a_i}} \\ \end{array}$$ $$n \sum_{i=1}^{n} \sqrt{\frac{a_i}{1 - a_i}} \leq \left(\sum_{i=1}^{n} a_i\right) \left(\sum_{i=1}^{n} \sqrt{\frac{1}{a_i(1 - a_i)}}\right)$$

The last inequality is true according to Chebyshev's inequality applied to the sequences

$$(a _ {1}, a _ {2}, \dots , a _ {n}) \text{ and } \left(\frac {1}{\sqrt {a _ {1} (1 - a _ {1})}}, \frac {1}{\sqrt {a _ {2} (1 - a _ {2})}}, \dots , \frac {1}{\sqrt {a _ {n} (1 - a _ {n})}}\right)$$

Theorem (Chebyshev's inequality) Let $a_1 \leq a_2 \leq \ldots \leq a_n$ and $b_1 \leq b_2 \leq \ldots \leq b_n$ be real numbers. Then we have

$$\left(\sum_ {i = 1} ^ {n} a _ {i}\right) \left(\sum_ {i = 1} ^ {n} b _ {i}\right) \leq n \left(\sum_ {i = 1} ^ {n} a _ {i} b _ {i}\right)$$

Note Chebyshev's inequality is also true if $a_1 \geq a_2 \geq \ldots \geq a_n$ and $b_1 \geq b_2 \geq \ldots \geq b_n$ .

But if $a_1 \leq a_2 \leq \ldots \leq a_n$ and $b_1 \geq b_2 \geq \ldots \geq b_n$ (or the reverse) then we have

$$\left(\sum_ {i = 1} ^ {n} a _ {i}\right) \left(\sum_ {i = 1} ^ {n} b _ {i}\right) \geq n \left(\sum_ {i = 1} ^ {n} a _ {i} b _ {i}\right)$$

$\frac{1}{1 + x_i} = a_i,$ for $i = 1,2,\ldots ,n,0\leq x_1\leq x_2\leq \ldots \ldots \leq x_n,n\geq 2,$ then

$$\frac {1}{1 + x _ {1}} \geq \frac {1}{1 + x _ {2}} \geq \dots \geq \frac {1}{1 + x _ {n}}$$ $$a _ {1} \geq a _ {2} \geq \dots \dots \geq a _ {n}$$

The sequence $(a_{1}, a_{2}, \ldots, a_{n})$ is non-increasing.

$$\frac {1}{\sqrt {a _ {i} (1 - a _ {i})}} = \frac {1}{\sqrt {\frac {1}{1 + x _ {i}} \left(1 - \frac {1}{1 + x _ {i}}\right)}} = \frac {1 + x _ {i}}{\sqrt {x _ {i}}}$$ $$f (x) = \frac {1 + x}{\sqrt {x}} = x ^ {- 1 / 2} + x ^ {1 / 2}, x > 0$$ $$f ^ {\prime} (x) = - \frac {1}{2} x ^ {- 3 / 2} + \frac {1}{3} x ^ {- 1 / 2} = \frac {x - 1}{2 x ^ {3 / 2}}$$

The function $f$ is increasing on $(1, \infty)$ .

$$f \left(\frac {1}{x}\right) = \frac {1 + \frac {1}{x}}{\sqrt {\frac {1}{x}}} = \frac {1 + x}{\sqrt {x}} = f (x)$$

If $x_{1} = 0$ , then $x_{2} \geq 0$

$$\frac {1}{1 + x _ {1}} + \frac {1}{1 + x _ {2}} \leq 1 \Rightarrow \frac {1}{1 + 0} + \frac {1}{1 + x _ {2}} \leq 1 \Rightarrow \frac {1}{1 + x _ {2}} \leq 0$$

This is false, if $x_{2} \geq 0$ .

If $0 < x_{1} < 1$ , then

$$\frac {1}{1 + x _ {1}} + \frac {1}{1 + x _ {2}} \leq 1$$ $$\frac {1}{1 + x _ {2}} \leq 1 - \frac {1}{1 + x _ {1}} = \frac {x _ {1}}{1 + x _ {1}} = > x _ {2} \geq \frac {1}{x _ {1}} = > x _ {2} > 1$$

Hence

$$f (x _ {2}) \leq \dots \leq f (x _ {n})$$

We have that

$$f (x _ {2}) \geq f \left(\frac {1}{x _ {1}}\right) = f (x _ {1})$$ $$f (x _ {1}) \leq f (x _ {2}) \leq \dots \leq f (x _ {n})$$

If $x_{1}\geq 1$, then

$$f (x _ {1}) \leq f (x _ {2}) \leq \dots \leq f (x _ {n})$$

It is clear that (both in case $x_{1} \geq 1$ and in case $x_{1} < 1$)

$$f (x _ {1}) \leq f (x _ {2}) \leq \dots \leq f (x _ {n})$$ $$\frac {1}{\sqrt {a _ {1} (1 - a _ {1})}} \leq \frac {1}{\sqrt {a _ {2} (1 - a _ {2})}} \leq \dots \leq \frac {1}{\sqrt {a _ {n} (1 - a _ {n})}}$$

The sequence $\left(\frac{1}{\sqrt{a_1(1 - a_1)}},\frac{1}{\sqrt{a_2(1 - a_2)}},\dots ,\frac{1}{\sqrt{a_n(1 - a_n)}}\right)$ is

non-decreasing.

Apply the Chebyshev's inequality

$$n \sum_ {i = 1} ^ {n} a _ {i} \sqrt {\frac {1}{a _ {i} (1 - a _ {i})}} \leq \left(\sum_ {i = 1} ^ {n} a _ {i}\right) \left(\sum_ {i = 1} ^ {n} \sqrt {\frac {1}{a _ {i} (1 - a _ {i})}}\right)$$ $$n \sum_ {i = 1} ^ {n} \sqrt {\frac {a _ {i}}{1 - a _ {i}}} \leq \left(\sum_ {i = 1} ^ {n} a _ {i}\right) \left(\sum_ {i = 1} ^ {n} \sqrt {\frac {1}{a _ {i} (1 - a _ {i})}}\right)$$

Therefore

$$\sqrt {x _ {1}} + \sqrt {x _ {2}} + \dots + \sqrt {x _ {n}} \geq (n - 1) \left(1 / \sqrt {x _ {1}} + \dots + 1 / \sqrt {x _ {n}}\right).$$

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