Prove that $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq 1$ Let $abcd=1$ and $a,b,c$ and $d$ are all positive.
Prove that
$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}+\dfrac{1}{(1+c)^2}+\dfrac{1}{(1+d)^2}\geq 1$
I am probably able to do this by assuming $a\geq b\geq c\geq d$
and by using derivative show that
$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\geq \dfrac{2}{(1+\sqrt{ab})^2}$
In this way I reduce the number of variables.
However this approach seems to be tedious.
Is there any other methods of proving this inequality?
I hope to learn the proof of the inequality in the title.

This inequality I have nice methods:use 
$$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\ge\dfrac{1}{1+ab}$$
pf:use $Cauchy-Schwarz$ inequality
we have
$$(a+1)^2\le\left(\dfrac{a}{b}+1\right)(ab+1)$$
$$(b+1)^2\le\left(\dfrac{b}{a}+1\right)(ab+1)$$
so
$$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\ge\dfrac{1}{\left(\dfrac{a}{b}+1\right)(ab+1)}+\dfrac{1}{\left(\dfrac{b}{a}+1\right)(ab+1)}=\dfrac{1}{ab+1}$$
$$\dfrac{1}{(1+c)^2}+\dfrac{1}{(1+d)^2}\ge\dfrac{1}{1+cd}$$
since $abcd=1$ then we have
$$\dfrac{1}{1+ab}+\dfrac{1}{1+cd}=\dfrac{1}{1+\dfrac{1}{cd}}+\dfrac{1}{1+cd}=\dfrac{cd}{1+cd}+\dfrac{1}{1+cd}=1$$
