## x,y,z>1

For discussing Olympiad Level Algebra (and Inequality) problems
photon
Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka
Contact:

### x,y,z>1

Prove that if $x,y,z$ $>$ $1$ , and $\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2$ , then
$\displaystyle \sqrt{x+y+z}\geq\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.$

Source - IMO longlist (1992)
Try not to become a man of success but rather to become a man of value.-Albert Einstein

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: x,y,z>1

As $\sum \frac{1}{x}=2$,$\sum \frac{x-1}{x}=1$
By Cauchy Schwarz inequality
$\sum x=\sum x \times \sum \frac{x-1}{x}\geq (\sum \sqrt{x-1})^{2}$
or $\sum x\geq (\sum \sqrt{x-1})^{2}$
or $\sqrt{\sum x}\geq \sum \sqrt{x-1}$
So $\sqrt{x+y+z}\geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$

photon
Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka
Contact:

### Re: x,y,z>1

that was cool

Let $x=a^2+1 , y=b^2+1 ,z=c^2+1$, ($a,b,c>0$)
then inequality becomes $\displaystyle \sqrt{a^2+b^2+c^2+3} \geq a+b+c$
$\Rightarrow a^2+b^2+c^2+3 \geq a^2+b^2+c^2+2ab+2bc+2ca$
$\Rightarrow \frac{3}{2}\geq ab+bc+ca$
We have $\sum \frac{1}{x}=2 \Rightarrow \sum \frac{1}{a^2+1}=2$
Let $f(x)=\frac{1}{x^2+1}$ , $f''(x)>0$ , by jenson's inequality ,
$\displaystyle \frac{1}{3} \sum \frac{1}{a^2+1} \le \frac{1}{(\frac{a+b+c}{3})^2+1}$
$\displaystyle \Rightarrow \frac{2}{3} \le \frac{9}{(a+b+c)^2+9}$
$\Rightarrow (a+b+c)^2 \le \frac{9}{2}$
$\Rightarrow \sum a^2 + 2\sum ab \le \frac{9}{2}$ ......(1)
Again , $f(x)=\frac{1}{x}$ , $f''(x)>0$
Again applying Jenson , $\frac {1}{3}[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}] \le \frac{3}{x+y+z}$
$\displaystyle \Rightarrow \frac{2}{3} \le \frac{3}{a^2+b^2+c^2+3}$
$\Rightarrow a^2+b^2+c^2 \le \frac{3}{2}$ .......(2)
$2(\sum a^2 + \sum ab) \le 6$
$\Rightarrow \sum a^2 + \sum ab \le 3$
as , $\sum ab \le \sum a^2$ - adding last 2 inequalities -
$2\sum ab \le 3 \Rightarrow ab+bc+ca \le \frac{3}{2}$ which satisfies the inequality .