Hard inequality :$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$ I have a hard problem this is it :

Let $a,b,c>0$ such that $a^ab^bc^c=1$ then we have :
  $$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$$

I try to use Jensen's ienquality applied to the function $f(x)=\frac{1}{x^2}$ but it doesn't works for all the values .
if we apply Am-Gm it's like Jensen's inequality so I forget this ways .
Maybe If we apply Karamata's inequality but I didn't found the right majorization .
I try also to use Muirhead inequality but without success .
So I'm a bit lost with that if you have a hint or a full answer I will be happy to read your work . 

First, I will ignore the constraint $a^ab^bc^c=1$ to sketch a possible solution for the aforementioned inequality, under an alternative constraint:
Set $x=\frac{1}{a^2+b^2}$ and $x=\frac{1}{b^2+c^2}$ and $z=\frac{1}{c^2+a^2}$.
From the aritmetic-geometric inequality
$$
x^2+y^2+z^2\geq 3(x^2y^2z^2)^{\frac{1}{3}},
$$
it remains to show that $$(x^2y^2z^2)^{\frac{1}{3}}\geq \frac{1}{4},$$
By noting that
$$ (a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2=\frac{1}{x^2y^2z^2},$$ the previous inequality to be shown becomes
$$
(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2 \leq 64.
$$
or equivalently
$$ (a^2+b^2)(b^2+c^2)(c^2+a^2)\leq 8,$$
if we take the square on both sides of the inequality.
Noteworty, if $(a,b),(b,c)$ and $(a,c)$ belong to the closure of the disc $\mathbb{T}$ of radius $\sqrt{2}$:
$$ \overline{\mathbb{T}}=\{(u,v)\in \mathbb{R}^2~:~\sqrt{u^2+v^2}\leq \sqrt{2}\},$$
the previous inequality is always satisfied.
