<s>[INST] Prove this Complicated Inequality 
Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \le 4$. Prove that $$\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.$$

Let $x = a + b, y = b + c, t = a + c$ Then the INEQ becomes, 
$$\frac{ab + 1}{(x)^2} + \frac{bc + 1}{(y)^2} + \frac{ca + 1}{(t)^2} \ge 3.$$
Any further hints?
 [/INST] $x = a+b,y= b+c,z = c+a$, then
$$x^2 + y^2 + z^2 \le 4$$
problem becomes
$$\sum\frac{4 + (x+z-y)(x+y-z)}{4x^2}=\sum \frac{4 + x^2 - y^2 - z^2 + 2yz}{4x^2}$$   
$$=\sum\frac{4-x^2 - y^2 -z^2 + 2x^2 + 2yz}{4x^2}\ge \sum\frac{2x^2 + 2yz}{4x^2} $$   
$$=\sum \left(\frac{1}{2} + \frac{1}{2}\frac{yz}{x^2}\right)\ge \frac{3}{2} +\frac{1}{2}\cdot 3\left(\prod\frac{yz }{x^2}\right)^{1/3} = 3$$
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