# A tricky algebraic inequality

This is an old inequality but I haven't seen a satisfactory solution yet and am hoping someone here can provide one. There are a couple of brute force solutions but they provide no insight into the inequality and I'd be surprised if there isn't a trick to it: $$x\frac{(y+z)^2}{(1+yz)^2} + y\frac{(x+z)^2}{(1+xz)^2} + z\frac{(x+y)^2}{(1+xy)^2} \ge \frac{3\sqrt 3}{4}$$ for $xy+yz+zx = 1$, all positive.

My attempt was to try to find a lower bound of the left side in terms of symmetric quantities like $u=x+y+z$ and $w=xyz$, however I haven't had much success despite multiple attempts. For example, the left hand side above can be bounded from below by $$\sum x\frac{(y+z)^2}{(1+yz)^2} \ge\frac{\left(\sum x(y+z)^2\right)^3}{\left(\sum (y+z)(1-y^2z^2)\right)^2} = \frac{\left(u+3w\right)^3}{\left(u+2u^2w+w\right)^2}$$ but the above has a minimum just a tad less than $3\sqrt 3/4$.

I have also tried a cotangent substitution and breaking the symmetry (i.e. assuming $x\ge y\ge z$) but I didn't get far.

I don't know the origin of the inequality but it's supposed to be competition level so I suspect it has a nice and tricky solution and not just a brute force one. So, I am hoping someone here will find it.

We need to prove that $$\sum_{cyc}\frac{x(y+z)^2}{(xy+xz+2yz)^2}\geq\frac{3}4\sqrt{\frac{3}{xy+xz+yz}}$$ or $$\sum_{cyc}\frac{\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)^2}{\left(\frac{1}{xy}+\frac{1}{xz}+\frac{2}{yz}\right)^2}\geq\frac{3}{4}\sqrt{\frac{3}{\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}}}$$ or $$\sum_{cyc}\frac{x(y+z)^2}{(y+z+2x)^2}\geq\frac{3}{4}\sqrt{\frac{3xyz}{x+y+z}}.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3.$

Thus, we need to prove that $$\frac{u(-11w^6+756u^3w^3-126uv^2w^3+972u^4v^2-576u^2v^4)}{(w^3+9uv^2+54u^3)^2}\geq\frac{1}{4}\sqrt{\frac{w^3}{u}}$$ or $f(v^2)\geq0$, where $$f(v^2)=4u(-11w^6+756u^3w^3-126uv^2w^3+972u^4v^2-576u^2v^4)-$$ $$-(w^3+9uv^2+54u^3)^2\sqrt{\frac{w^3}{u}}.$$ Now, it's obvious that $f''(v^2)<0$, which says that $f$ is a concave function.

But the concave function gets a minimal value for an extreme value of $v^2$,

which happens for equality case of two variables.

Since $f(v^2)\geq0$ is a homogeneous inequality it's enough to assume that $y=z=1$

and we need to prove that $$\frac{4x}{(2+2x)^2}+\frac{2(x+1)^2}{(x+3)^2}\geq\frac{3}{4}\sqrt{\frac{3x}{x+2}}$$ or $$(x-1)^2(37x^7+346x^6+1339x^5+2700x^4+2891x^3+1466x^2+309x+128)\geq0.$$ Done!

• @dang dang haj It just turned out. – Michael Rozenberg May 24 '18 at 14:14
• @dang dang haj Do you ask me, how I got this? – Michael Rozenberg May 25 '18 at 3:38