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$$ but it's not. How to prove it?
Best Answer
Observations towards a solution
As suggested by Siddharth, show that $ x = \frac{ -a^2 + b^2 + c^2 } { 2bc }$.
As remarked by Abastro, $x, y, z$ are cosines of angles of the a-b-c triangle. In particular $x, y, z \leq 1$ (though we won't need this).
In fact, the inequality holds for any triangle (with positive sides), without the condition that $ x, y, z \geq 0$. The proof still works, but you just have to keep track of signs (See point 6.)
If we allow degenerate triangles, observe that $ (x,y, z) = (0, 0, 1)$ (and permutations), yield the equality condition. This implies that a simple AM-GM/Jensens, etc will likely not work, esp if it only yields the $ x = y = z = 1/2$ equality condition.
We likely need to use methods like Schur's inequality, Mixing Variable, Vasc's RCF theorem, etc, to get to the other equality case.
The (naive) direct application of Titu's lemma yields
$$ \sum \frac{ x^2 } { 1 + x} \geq \frac{ ( x+y+z ) ^2 } { 3 + x + y + z}.$$
However, with $ (x, y, z ) \rightarrow (0,0,1)$, the RHS $\rightarrow \frac{1}{4}$, so we've over-optimized. We need to figure out how to weight these terms, which isn't that obvious.
$$ \sum \frac{ x^2 } { 1+x } \geq \frac{ ( a^2 +b^2 + c^2) ^2 } {\sum 2bc( -a^2 + b^2 + c^2 + 2bc) }. $$
Notes for this step:
Hence
$$ \sum \frac{ x^2 } { 1+x } \geq \sum \frac{ ( a^2 +b^2 + c^2) ^2 } {\sum 2bc( -a^2 + b^2 + c^2 + 2bc) } \geq \frac{1}{2}. $$