Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $

Question

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?

Answer 1

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)

Answer 2

Observations towards a solution

1. As suggested by Siddharth, show that $ x = \frac{ -a^2 + b^2 + c^2 } { 2bc }$.

2. 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).

3. 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.)

4. 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.

5. 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.

6. 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 $.

7. 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

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

8. 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 $(x,y,z) = (0, 0, 1)$.