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 . 
 A: Define: $$F(a,b,c)=\left(\frac{1}{a^2+b^2}\right)^2+\left(\frac{1}{b^2+c^2}\right)^2+\left(\frac{1}{c^2+a^2}\right)^2$$
By the method of Lagrange multipliers, the minimizer of $F$ under the constraint $a^ab^bc^c=1$ must be a critical point of the Lagrangian
$$L(a,b,c,\lambda)=\left(\frac{1}{a^2+b^2}\right)^2+\left(\frac{1}{b^2+c^2}\right)^2+\left(\frac{1}{c^2+a^2}\right)^2+\lambda(a^ab^bc^c-1)$$
To find the critical points we solve $\nabla L=0$:
$$\frac{\partial L}{\partial\lambda}=a^ab^bc^c-1=0$$
$$\frac{\partial L}{\partial a}=-\frac{4a}{(a^2+b^2)^3}-\frac{4a}{(a^2+c^2)^3}+\lambda(a^ab^bc^c)(1+\log a)=0$$
and similarly find the equation for $\frac{\partial L}{\partial b}=0$ and $\frac{\partial L}{\partial c}=0$. We got a system of 4 (non-linear) equations with 4 variables.
Let $a,b,c,\lambda$ be a solution. Since $a^ab^bc^c=1$, at least one of $a,b,c$ is $\leq 1$ and at least one is $\geq 1$, so WLOG we can treat two cases: $a\leq b\leq 1\leq c$ and $a\leq 1\leq b\leq c$ .


*

*First case $a\leq b\leq 1\leq c$: Isolating $\lambda$ in the equations $\frac{\partial L}{\partial a}=0$ and $\frac{\partial L}{\partial b}=0$ leads to
$$\lambda=\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(a^2+c^2)^3}\right]\cdot\frac{a}{1+\log a}=\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(b^2+c^2)^3}\right]\cdot\frac{b}{1+\log b}$$
If $a<b$ then 
$$\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(a^2+c^2)^3}\right]>\left[\frac{1}{(a^2+b^2)^3}+\frac{1}{(b^2+c^2)^3}\right]$$
and 
$$\frac{a}{1+\log a}>\frac{b}{1+\log b}$$
by contradiction to the equality above. Therefore we conclude $a=b$, and we may reduce our problem to 2 dimensions: Minimize
$$F(a,a,c)=\frac{1}{4a^4}+2\left(\frac{1}{a^2+c^2}\right)^2$$
Among all $a,c$ s.t. $a^{2a}c^c=1$. Using the square super root function we put $c=\text{ssrt}(a^{-2a})$ in the above expression and get
$$G(a)=\frac{1}{4a^4}+2\left(\frac{1}{a^2+\text{ssrt}(a^{-2a})^2}\right)^2$$
Convince yourself that $G$ attains its minimum at $a=1$ (I used a computer, maybe possible to do it analytically), so to minimize $G$ we must take $a=1$. We conclude $a=b=1$, and then $c=\text{ssrt}(1)=1$ as well.

*Second case $a\leq 1\leq b\leq c$: Isolating $\lambda$ in the equations $\frac{\partial L}{\partial b}=0$ and $\frac{\partial L}{\partial c}=0$ leads to
$$\lambda=\left[\frac{1}{(b^2+a^2)^3}+\frac{1}{(b^2+c^2)^3}\right]\cdot\frac{b}{1+\log b}=\left[\frac{1}{(c^2+a^2)^3}+\frac{1}{(c^2+b^2)^3}\right]\cdot\frac{c}{1+\log c}$$
If $b<c$ then 
$$\frac{1}{(b^2+a^2)^3}\frac{b}{1+\log b}>\frac{1}{(c^2+a^2)^3}\frac{c}{1+\log c}$$
and 
$$\frac{1}{(b^2+c^2)^3}\frac{b}{1+\log b}>\frac{1}{(c^2+b^2)^3}\frac{c}{1+\log c}$$
by contradiction to the equality above. Therefore we conclude $b=c$, and continue as in the first case to show $a=b=c=1$.
To conclude: 
$$\min\{F(a,b,c);a^ab^bc^c=1\}=F(1,1,1)=\frac{3}{4}$$
Remark: I may have over-complicated things. If you can show in a more simple way that $a=b=c=\lambda=1$ is the only solution to $\nabla L=0$ then you are done.
A: 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.
A: Hint.
Making $b = \lambda a, c = \mu a$ we have
$$
\frac{1}{(1+\lambda^2)^2}+\frac{1}{(\lambda^2+\mu^2)^2}+\frac{1}{(\mu^2+1)^2}\ge \frac 34 a^4
$$
and
$$
a = \frac{1}{(\mu^{\mu}\lambda^{\lambda})^{\frac{1}{1+\lambda+\mu}}}
$$
then if 
$$
f(\lambda,\mu) = \frac{1}{(1+\lambda^2)^2}+\frac{1}{(\lambda^2+\mu^2)^2}+\frac{1}{(\mu^2+1)^2}- \frac 34 \frac{1}{(\mu^{\mu}\lambda^{\lambda})^{\frac{4}{1+\lambda+\mu}}}
$$
now we need to demonstrate that 
$$
\min f_{\lambda,\mu} \ge 0
$$
Note that 
$$
f(\lambda,\lambda) \le f(\lambda,\mu)
$$
It is easy to verify also that along $\lambda = \mu$
$$
f(1,1) = 0\\
f'(1,1) = 0\\
f''(1,1) > 0
$$
