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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.

2 Answers 2

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$p=a+b+c$ and $q=ab+bc+ca$. Using $AM-GM$ we obtain that : $p,q \geq 3.$ The inequality is equivalent with:

$\dfrac{3+4p+q+p^2}{2p+q+p^2+pq} \leq \dfrac{12+4p+q}{9+4p+2q} \Leftrightarrow$ $3p^2q+pq^2+6pq-5p^2-q^2-24p-3q-27 \geq 0 \Leftrightarrow \\\left(3p^2q-5p^2-12p\right)+\left(pq^2-q^2-3p-3q\right)+\left(6pq-9p-27\right) \geq 0.$

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    @Unoqualunque: I didn't know that. Interesting! Thanks +12012-09-07
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$\frac{2}{1+a+b}-(\frac{1}{2+a}+\frac{1}{2+b})$

$=\frac{1}{1+a+b}-\frac{1}{2+a}+\frac{1}{1+a+b}-\frac{1}{2+b}$

$=\frac{1}{1+a+b}(\frac{1-b}{2+a}+\frac{1-a}{2+b})$

$=\frac{1}{(1+a+b)(2+a)(2+b)}((1-b)(2+b)+(1-a)(2+a))$

$≤\frac{1}{1\cdot 2\cdot 2}(4-(a+b+a^2+b^2))$ 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 $n$.

$\implies \sum2(\frac{1}{1+a+b})≤ \sum2(\frac{1}{2+a})$

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    Could you please let me know how to block or unblock?2012-09-09