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?
 A: 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.


*As suggested by Siddharth, using the substitution $ x = \frac{ -a^2 + b^2 + c^2 } { 2bc }$, we get $ \frac{ x^2 } { 1 + x }  = \frac{ (-a^2 + b^2 + c^2 ) ^2 } { 2bc( -a^2 + b^2 + c^2  + 2bc)} $. Now, applying Titu's lemma gives us

$$ \sum \frac{ x^2 } { 1+x } \geq \frac{ ( a^2  +b^2 + c^2) ^2 } {\sum 2bc( -a^2 + b^2 + c^2  + 2bc) }. $$
Notes for this step:

*

*By clearing denominators, we've actually weighted the terms. As it turns out in the rest of the solution, we got lucky.

*Work through this if you want a solution to "all triangles" instead of just "acute triangles". While it seems like I'm using that $ x\geq 0 \Rightarrow b^2 +  c^2 \geq a^2 $ in this part of the proof so that we don't flip the signs in the numerator, that isn't necessary. In the case of $ x < 0 \Rightarrow b^2 + c^2 < a^2$, I can still apply Titu as above because $1 + x > 0 $ and $ -x > x $.



*Now, observe that $ ( a^2  +b^2 + c^2) ^2  \geq \sum bc( -a^2 + b^2 + c^2  + 2bc)$ when expanded out is just the Schur's inequality $ \sum a^2 ( a-b)(a-c) \geq 0 $. (Verify this yourself, I'm too lazy to show the steps.)
Hence

$$ \sum \frac{ x^2 } { 1+x } \geq \sum \frac{ ( a^2  +b^2 + c^2) ^2 } {\sum 2bc( -a^2 + b^2 + c^2  + 2bc) } \geq \frac{1}{2}. $$


*The equality case is when


*

*All values are equal: $ a=b= c \Rightarrow x = y = z = \cos 60^\circ = 1/2$.

*Two of them are equal and the other is zero: $a=b, c =0$ is the degenerate isosceles triangle, which yields $(x,y,z) = (0, 0, 1)$.

A: 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)

