# 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 $$cy+bz =a; az + cx = bbx + ay = c$$. 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?

• Try calculating value of $a$ in terms of $x,y,z$ and then apply AM-GM.
– user1135351
Jan 11 at 8:06
• Alternatively, you can observe that x, y, z are cosines of angles of the abc - triangle. Jan 11 at 8:19
• @Abastro I had tried doing that, but no result. Jan 11 at 13:39
• It is often useful to find the equality condition, which could suggest which methods work (and more importantly, which methods are likely not to). $\quad$ In this case, why would a simplistic Jensen's not work? What is the equality condition of your Jensen's? Jan 11 at 15:37
• Here is a solution. Part of it is not my work so I will comment but not answer. First, eliminate $c$ by the first two equations to get $a:b=\frac1{x^2+yz}:\frac1{y^2+zx}$, so$$a:b:c=\frac1{x^2+yz}:\frac1{y^2+zx}:\frac1{z^2+yx}.$$Plug ratio in any equation to get $x^2+y^2+z^2+2xyz=1$. Then just look at artofproblemsolving.com/community/c6h59649p1291749 to finish off.
– user1034536
Jan 11 at 17:39

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.

1. 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 $$-x > x$$.
1. 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

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

1. 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 $$(x,y,z) = (0, 0, 1)$$.
• @Anonymous Yes, strongly encourage you to upload your own solutions too. The observation of the equality case was what my comments were hinting at (and sorry for the initial confusion). Bear in mind that you'd have to deal with this too for your solution. Jan 11 at 16:44
• I was going false. My inequality contradicted the original one. That's why deleted comment Jan 11 at 16:48
• May I ask what Titu's lemma is? Jan 11 at 16:50
• @CalvinLin I also have an solution. Quite similar approach till substituting $x=cosA$ and then using AM-GM. (Hope you don't mind that I'll be posting my approach as an answer later)
– user1135351
Jan 11 at 16:51
• @SaiMehta It's a version of Cauchy Schwarz, used for summing of fractions. See here for more info. Jan 11 at 16:52

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)

• @CalvinLin my function is actually $\cos A$ one. So I would give equality there which you already told. (Although for this equality holds when $u=v=w= \frac{1}{\sqrt{3}}$)
– user1135351
Jan 11 at 18:17
• Okay, this works now. Jan 11 at 18:22
• @CalvinLin do you meant to say. $x=y=z=1/2$ isn't a single case where equality holds?
– user1135351
Jan 11 at 18:27
• @CalvinLin and Siddharth I am now confused which answer should I accept. As I both are awesome and short. Jan 11 at 18:31
• For the stated AM-GM, you also have an equality condition of $u = 0$, since you're multiplying by $u^2$. $\quad$ Then, the case of $u = v = 0$ also satisfies all 3 inequalities, and this gives the $(x, y, z) \rightarrow (0, 0, 1)$ case. That modification explains how you introduced the other equality condition. Jan 11 at 18:32