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
I will post them if you want
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