# classic inequality

• Jan 4th 2008, 06:54 AM
DivideBy0
classic inequality
Prove that if x, y, and z are positive real numbers, then

$\displaystyle \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}$

I haven't done much with inequalities so please explain your steps, thanks!
• Jan 4th 2008, 09:47 AM
Isomorphism
Quote:

Originally Posted by DivideBy0
Prove that if x, y, and z are positive real numbers, then

$\displaystyle \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}$

I haven't done much with inequalities so please explain your steps, thanks!

This is Nesbitts inequality, I know many proofs for this. Nesbitt's inequality - Wikipedia, the free encyclopedia
The above link proves it using AM-GM and Rearrangement inequality
I know a few more too :D
I will post them if you want
• Jan 4th 2008, 10:43 AM
PaulRS
Define $\displaystyle x+y+z=\lambda>0$

THus: $\displaystyle y+z=\lambda-x$; $\displaystyle x+z=\lambda-y$; $\displaystyle x+y=\lambda-z$

So we have to prove: $\displaystyle \frac{x}{\lambda-x}+\frac{y}{\lambda-y}+\frac{z}{\lambda-z}\geq{\frac{3}{2}}$

Let: $\displaystyle f(x)=\frac{x}{\lambda-x}=\frac{\lambda}{\lambda-x}-1$ which is clearly convex in $\displaystyle (0; \lambda)$ (use the second derivative)

It follows from Jensen's Inequality that: $\displaystyle \frac{1}{3}\cdot{\left(f(x)+f(y)+f(z)\right)}\geq{ f\left(\frac{x+y+z}{3}\right)}$

Therefore: $\displaystyle \frac{1}{3}\cdot{\left(\frac{x}{\lambda-x}+\frac{y}{\lambda-y}+\frac{z}{\lambda-z}\right)}\geq{\frac{\left(\frac{x+y+z}{3}\right)} {\lambda-\left(\frac{x+y+z}{3}\right)}}$

Now remember that: $\displaystyle x+y+z=\lambda$ and the assertion follows

Jensen's inequality - Wikipedia, the free encyclopedia
• Jan 4th 2008, 03:48 PM
DivideBy0
Thanks!
• Jan 5th 2008, 07:12 PM
ThePerfectHacker
It can also be done with AM-HM, that is how I learned it.