How may I prove this inequality? Let $a, b, c$ be positive real, $abc = 1$. Prove that:
$$\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} \le \frac{1}{2+a} + \frac{1}{2+b}+\frac{1}{2+c}$$
I thought of Cauchy and AM-GM, but I don't see how to successfully use them to prove the inequality. Any hint, suggestion will be welcome. Thanks.

$\frac{2}{1+a+b}-(\frac{1}{2+a}+\frac{1}{2+b})$
$=\frac{1}{1+a+b}-\frac{1}{2+a}+\frac{1}{1+a+b}-\frac{1}{2+b}$
$=\frac{1}{1+a+b}(\frac{1-b}{2+a}+\frac{1-a}{2+b})$
$=\frac{1}{(1+a+b)(2+a)(2+b)}((1-b)(2+b)+(1-a)(2+a))$
$≤\frac{1}{1\cdot 2\cdot 2}(4-(a+b+a^2+b^2))$ as $a,b>0,2+a>2$ and $a+b+1>1$
$\sum(\frac{2}{1+a+b}-(\frac{1}{2+a}+\frac{1}{2+b}))≤\frac{1}{4}(3\cdot 4-2(a+b+c)-2(a^2+b^2+c^2))≤0$ as $a^n+b^n+c^n≥3(abc)^{\frac{n}{3}}=3$ for any positive number $n$.
$\implies \sum2(\frac{1}{1+a+b})≤ \sum2(\frac{1}{2+a})$
