<s>[INST] Prove that $\sum\limits_{cyc}\left(\frac{a+b}{a+b+c}\right)^2\geq\frac{16}{9}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\left(\frac{a+b}{a+b+c}\right)^2+\left(\frac{b+c}{b+c+d}\right)^2+\left(\frac{c+d}{c+d+a}\right)^2+\left(\frac{d+a}{d+a+b}\right)^2\geq\frac{16}{9}$$
I tried C-S and more, but without success. 
I am looking for an human proof, which we can use during competition. 
 [/INST] HINT
Using substitutions $x =a+b+c, y = b+c+d, z = c+d+a, t = a+b+d$ we get $$(\frac{a+b}{a+b+c})^2=\frac 1 9(2 + \frac t x - \frac y x -\frac z x)^2$$
Similar for the other terms. Eliminate the squares then use the fact that $r + \frac 1 r \ge 2, r \gt 0$. 
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