<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] By AM-GM
$$\sum_{cyc}\frac{ab+1}{(a+b)^2}=\frac{1}{2}\sum_{cyc}\frac{2ab+2}{(a+b)^2}\geq\frac{1}{2}\sum_{cyc}\frac{2ab+a^2+b^2+c^2+ab+ac+bc}{(a+b)^2}=$$
$$=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{c^2+ab+ac+bc}{(a+b)^2}=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}\geq$$
$$\geq\frac{3}{2}+\frac{1}{2}\cdot3\sqrt[3]{\prod_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}}=\frac{3}{2}+\frac{3}{2}=3.$$
Done!
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