Let $x,y,z$ be positive real numbers such that $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$.
Find the minimum value of $$\frac{x^2y^2} {z^3(x^2+y^2)}+\frac {y^2z^2} {x^3(y^2+z^2)}+\frac {z^2x^2} {y^3(z^2+x^2)}$$ I'm pretty sure that the answer would be $\frac {\sqrt {3}} {2}$, when all parameters are $\sqrt {3}$. But I couldn't prove it after some hours of thinking. So can anyone help me? Any help would be welcome. Thanks:D.
Let $$a=\dfrac{1}{x},b=\dfrac{1}{y},c=\dfrac{1}{z},a^2+b^2+c^2\ge 1$$ Use Cauchy-Schwarz $$\sum_{cyc}\dfrac{x^2y^2}{z^3(x^2+y^2)}=\sum_{cyc}\dfrac{c^3}{(a^2+b^2)}\ge \dfrac{(a^2+b^2+c^2)^2}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}$$ since $$c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)\le\dfrac{2}{3}(a+b+c)(a^2+b^2+c^2)\le\dfrac{2}{\sqrt{3}}(a^2+b^2+c^2)^{\frac{3}{2}}$$ so $$LHS\ge \dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}\ge\dfrac{\sqrt{3}}{2}$$
If $x=y=z=\sqrt3$ then we get a value $\frac{\sqrt3}{2}$.
We'll prove that it's a minimal value, for which it's enough to prove that $$\sum_{cyc}\frac{y^2z^2}{x^3(y^2+z^2)}\geq\frac{\sqrt3}{2}\cdot\sqrt{\frac{x^2y^2+x^2z^2+y^2z^2}{x^2y^2z^2}}$$ or
$$\sum_{cyc}\frac{y^3z^3}{x^2(y^2+z^2)}\geq\frac{\sqrt3}{2}\cdot\sqrt{x^2y^2+x^2z^2+y^2z^2}.$$ Let $xy=c$, $xz=b$ and $yz=a$ and since our inequality is homogeneous,
we can assume that $a^2+b^2+c^2=3$.
Thus, we need to prove that $$\sum_{cyc}\frac{a^3}{3-a^2}\geq\frac{3}{2}$$ or $$\sum_{cyc}\left(\frac{a^3}{3-a^2}-\frac{1}{2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)(2a^2+3a+3)}{3-a^2}\geq0$$ or $$\sum_{cyc}\left(\frac{(a-1)(2a^2+3a+3)}{3-a^2}-2(a^2-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2(2a^2+6a+3)}{3-a^2}\geq0.$$ Done!
