A tricky algebraic inequality

Question

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.

Answer 1

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!

Answer 2

Another way: Let $xy=c$, $xz=b$ and $yz=a$.

Hence, $a+b+c=1$ and we need to prove that $$\sum_{cyc}\frac{a(b+c)^2}{(2a+b+c)^2}\geq3\sqrt{\frac{3abc}{16(a+b+c)}}$$ and since the last inequality is homogeneous, we can forgot the condition $a+b+c=1$.

Now, by AM-GM $$\sum_{cyc}\frac{a(b+c)^2}{(2a+b+c)^2}=\sum_{cyc}\frac{144a(b+c)^2(2a+5b+5c)^2}{\left(2\sqrt{(6a+3b+3c)(2a+5b+5c)}\right)^4}\geq$$ $$\geq\sum_{cyc}\frac{144a(b+c)^2(2a+5b+5c)^2}{\left(6a+3b+3c+2a+5b+5c\right)^4}=\sum_{cyc}\frac{9a(b+c)^2(2a+5b+5c)^2}{256(a+b+c)^4}.$$ Thus, it's enough to prove that $$3\sum_{cyc}a(b+c)^2(2a+5b+5c)^2\geq64(a+b+c)^3\sqrt{3abc(a+b+c)}$$ or $$3\sum_{cyc}a(b+c)^2(2a+5b+5c)^2-64(a+b+c)^3(ab+ac+bc)+$$ $$+64(a+b+c)^3\left(ab+ac+bc-\sqrt{3abc(a+b+c)}\right)\geq0$$ or $$3\sum_{cyc}a(b+c)^2(2a+5b+5c)^2-64(a+b+c)^3(ab+ac+bc)+$$ $$+\frac{64(a+b+c)^3\left((ab+ac+bc)^2-3abc(a+b+c)\right)}{ab+ac+bc+\sqrt{3abc(a+b+c)}}\geq0$$ or $$3\sum_{cyc}a(b+c)^2(2a+5b+5c)^2-64(a+b+c)^3(ab+ac+bc)+$$ $$+\frac{32(a+b+c)^3\sum\limits_{cyc}a^2(a-b)^2}{ab+ac+bc+\sqrt{3abc(a+b+c)}}\geq0.$$ But $\sqrt{3abc(a+b+c)}\leq ab+ac+bc$.

Thus, it remains to prove that $$3\sum_{cyc}a(b+c)^2(2a+5b+5c)^2-64(a+b+c)^3(ab+ac+bc)+$$ $$+\frac{16(a+b+c)^3\sum\limits_{cyc}a^2(a-b)^2}{ab+ac+bc}\geq0.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, easy to see that the last inequality is a linear inequality of $w^3$,

which says that it's enough to prove the last inequality for an extremal value of $w^3$,

which happens in the following cases.

1. $b=c=1$, which gives $$(a-1)^2(38a^3+87a^2-36a+19)\geq0,$$ which is obvious;

2. $w^3=0$.

Let $c=0$ and $b=1$.

Here we obtain: $$a(43a^3-24a^2-24a+43)\geq0.$$ Done!