Inequality

**james_bond** · January 1st 2008, 07:49 AM

Prove that if $a;b\in\mathbb{R}^+$ and $a+b=1$ , then

$\left(a+\frac 1a\right)^2+\left(b+\frac 1b\right)^2\ge \frac {25}2$ .
Similarly to the previous problem:
Prove that if $a;b;c\in\mathbb{R}^+$ and $a+b+c=1$ , then

$\left(a+\frac 1a\right)^2+\left(b+\frac 1b\right)^2+\left(c+\frac 1c\right)^2\ge \frac {100}3$ .
Let's generalize the problems above:
Prove that if $a_1;a_2;\ldots ;a_n\in\mathbb{R}^+$ , $n\ge 2$ , $n\in\mathbb{N}$ , and $a_1+a_2+\ldots +a_n=1$ then

$\left(a_1+\frac 1{a_1}\right)^2+\left(a_2+\frac 1{a_2}\right)^2+\ldots +\left(a_n+\frac 1{a_n}\right)^2\ge K$ .
Find the maximum possible value of $K$ .

**Isomorphism** · January 1st 2008, 08:48 AM

Originally Posted by james_bond

Prove that if $a;b\in\mathbb{R}^+$ and $a+b=1$ , then

$\left(a+\frac 1a\right)^2+\left(b+\frac 1b\right)^2\ge \frac {25}2$ .

Equality for $a=b=\frac12$
By Cauchy-Schwarz we have $(a+\frac 1a)^2+(b+\frac 1b)^2 \geq \frac{(a+\frac 1a + b+\frac 1b)^2}{1 + 1} = \frac{(1+\frac{1}{ab})^2}{2}$

Since $(a+b)^2 = 1 \geq 4ab \Rightarrow \frac{(1 + \frac1{ab})^2}{2} \geq \frac{25}{2}$ .

Thus we have our solution

**Isomorphism** · January 1st 2008, 09:07 AM

I post the remaining separately because a small trick there, wont work any longer

I will directly attack the generalization,

Claim: $K = \frac{(n^2 + 1)^2}{n}$
Proof:
Denote LHS of your inequality by I,
AM-HM says:
$(\sum a_i)(\sum \frac1{a_i}) \geq n^2 \Rightarrow (1 + \sum \frac1{a_i})^2 \geq (n^2 + 1)^2$ -------------(*)
Applying CS:
$I \geq \frac{{(a_1+\frac 1{a_1} + a_2+\frac 1{a_2} + \ldots a_n+\frac 1{a_n}})^2}{1+1+1 \ldots +1}$
But $({a_1+\frac 1{a_1} + a_2+\frac 1{a_2} + \ldots a_n+\frac 1{a_n}} = 1 + \sum \frac1{a_i}$
So $I \geq \frac{{(1 + \sum \frac1{a_i})}^2}{n}$ .
But by (*), $\frac{{(1 + \sum \frac1{a_i})}^2}{n}\geq \frac{(n^2 + 1)^2}{n}$
Thus $I \geq \frac{(n^2 + 1)^2}{n}$
Equality for all $a_i$ s equal.
K cant get any better than that because we showed equality achieving

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