Q:

USA $2011$ contest inequality problem, proving $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3$, under given condition.

If $a^2+b^2+c^2+(a+b+c)^2\le4$, then $$\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3.$$

My attempt:
From the given criteria, one can easily obtain that $$(a+b)^2+(b+c)^2+(c+a)^2\le 4\tag{$\clubsuit$}$$
Now, I tried Cauchy-Scwartz on the sets $$\left\{\frac{\sqrt{ab+1}}{a+b},\frac{\sqrt{bc+1}}{b+c},\frac{\sqrt{ca+1}}{c+a}\right\}\text{and}\left\{\sqrt{ab+1}(a+b),\sqrt{bc+1}(b+c),\sqrt{ca+1}(c+a)\right\},$$
and got $$\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge\frac{(ab+bc+ca+3)^2}{\sum(ab+1)(a+b)^2}\\
\ge\frac{(ab+bc+ca+3)^2}{\sum ab(a+b)^2+4}[\text{see}\; (\clubsuit)\text{ marked inequality}]$$
Now, what is left is just proving $$\frac{(ab+bc+ca+3)^2}{\sum ab(a+b)^2+4}\ge 3,$$ i.e. (after some simple arithmetic), $$6ab+6bc+6ca+2a^bc+2ab^2c+2abc^2\ge3+a^3b+ab^3+b^3c+bc^3+c^3a+ca^3+a^2b^2+b^2c^2+c^2a^2.$$
Now, I got totally stuck! Even Muirhead fails to majorise the powers of LHS, so, any idea......
P.S. Though I am tagging this (algebra-precalculus), I won't mind, if someone uses calculus. Any kind of solutions are welcome, but still, I'd like precalculus more, since it is a contest problem. 

A:

For positives $a$, $b$ and $c$ by AM-GM we obtain:
$2\sum\limits_{cyc}\frac{1+ab}{(a+b)^2}\geq\sum\limits_{cyc}\frac{a^2+b^2+c^2+ab+ac+bc+2ab}{(a+b)^2}=3+\sum\limits_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}\geq3+3=6$

