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> Let $a$, $b$ and $c$ be three positives such that $abc=1$. Prove that $$\sum\_{cyc}\frac1{(b+1)^2}+\frac1{a+b+c+1}\ge1$$
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Here's what have I done that is completely incorrect. Let $a + b + c + 1 = x$. We have that
$$\displaystyle \sum\_{cyc}\dfrac{1}{(b + 1)^2} \ge \displaystyle \sum\_{cyc}\dfrac{1}{(c + 1)(a + 1)} = \sum\_{cyc}\dfrac{1}{x - b + ca}$$
That means
$$\displaystyle \sum\_{cyc}\dfrac{1}{(b + 1)^2} + \dfrac{1}{a + b + c + 1} \ge \dfrac{16}{bc + ca + ab + 3x + 1}$$
$$\ge \dfrac{48}{(x - 1)^2 + 9x + 3} = \dfrac{48}{x^2 + 7x + 4}$$
That was illogical.