Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $ 
Assume that positive numbers a, b, c, x, y, z satisfy $cy+bz =a;   
 az + cx = b$$bx + ay = c$.
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $

I've tried appling A.M.-G.M. inequality but it didn't work. Then applied jensen inequality assuming, $f(x)=\frac{x^2}{1+x}$
As the function is convex for $x>0$,
$$f\left(\frac{x+y+z}{3}\right)\leq \frac{f(x)+f(y)+f(z)}{3}$$
The thing we need to prove is $$f\left(\frac{x+y+z}{3}\right)=1/6$$ but it's not. How to prove it?

All things are same as that of answer given by Calvin Lin upto $x= \frac{b^2+c^2-a^2}{2bc}$ and similarly for $y,z$.
Clearly we can see that $b^2+c^2>a^2$ ($x$ is positive) and same for $b,c$.
We get an acute angled triangle of sides $a,b,c$ with $x=\cos A, y=\cos B, z=\cos C$. Let the function which to prove is $f(\Delta)$
Now we have to find minimum value of $$f(\Delta)= \frac{\cos^2 A}{1+\cos A}+\frac{\cos^2 B}{1+\cos B}+\frac{\cos^2 C}{1+\cos C}$$
For $A+B+C=\pi$
Now let $u=\cot A, v=\cot B, w=\cot C$
It can be easily proved that,
$$\frac{\cos^2 A}{1+\cos A}=u^2- \frac{u^3}{\sqrt{(u+v)(u+w)}} $$

Hint: Use $uv+vw+wu=1$ for a triangle.

By A.M.$\geq$G.M.,
$$(u+v)+(u+w) \geq 2 \sqrt{(u+v)(u+w)}$$
Or,
$$\frac{1}{u+w}+ \frac{1}{u+v} \geq \frac{2}{\sqrt{(u+v)(u+w)}} $$
Rewriting the above inequality in a different way gives,
$$u^2- \frac{u^3}{\sqrt{(u+v)(u+w)}} \geq u^2-\frac{u^3}{2}\left(\frac{1}{u+w}+ \frac{1}{u+v}\right)$$
Or,
$$\frac{\cos^2 A}{1+\cos A}\geq u^2-\frac{u^3}{2}\left(\frac{1}{u+v}+ \frac{1}{u+w}\right)$$
Adding this inequality for $\cos A, \cos B, \cos C$ gives,
$$f(\Delta)\geq \frac{uv+vw+wu}{2}= \frac{1}{2} $$

One value of $(u,v,w)$ for which to hold equality is $\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.( Although equality can hold at other places as well, as described by Calvin Lin)

