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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.

user 1591719
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2 Answers2

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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:

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

Iuli
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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 $n$.

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

lab bhattacharjee
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  • Can anybody please verify the following: as $\sum (4-(a^2+b^2+a+b))<0$ at least one of $4-(a^2+b^2+a+b)$ must be $<0$. In that case, does $\frac{1}{(1+a+b)(2+a)(2+b)}((1-b)(2+b)+(1-a)(2+a))≤\frac{1}{1\cdot 2\cdot 2}(4-(a+b+a^2+b^2))$ hold as the numerator $<0$ and denominator is decreased? Like, $-\frac{4}{9}<-\frac{4}{7}?$ – lab bhattacharjee Sep 08 '12 at 07:05
  • I think you blocked me to edit the questions because I want to improve your post putting "\displaystyle" before the equations. – Iuli Sep 08 '12 at 21:39
  • Could you please let me know how to block or unblock? – lab bhattacharjee Sep 09 '12 at 06:07