Prove the inequality $\left(\frac1a+\frac1b+\frac1c\right)\left(\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\right)\ge\frac9{1+abc}$

Let $$a,b,c>0$$. Prove the inequality $$\left(\frac1a+\frac1b+\frac1c\right)\left(\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\right)\ge\frac9{1+abc}$$

My work so far:

Use AM-GM: $$\frac1{1+a}+\frac1{1+b}+\frac1{1+c}=$$ $$=\frac{\frac1a}{1+\frac1a}+\frac{\frac1b}{1+\frac1b}+\frac{\frac1c}{1+\frac1c}\ge$$ $$\ge3\sqrt[3]{\frac{\frac1a\cdot\frac1b\cdot\frac1c}{\left(1+\frac1a\right)\left(1+\frac1b\right)\left(1+\frac1c\right)}}=$$ $$\ge\frac3{\sqrt[3]{abc}}\sqrt[3]{\frac{1}{\left(1+\frac1a\right)\left(1+\frac1b\right)\left(1+\frac1c\right)}}$$

• Where did you find this problem? As this is pretty clearly a contest-style inequality, what contest is this from? Commented Aug 30, 2016 at 17:58
• Using the HM-GM-AM inequality, we have $$\frac 3{HM} = \frac 1a + \frac 1b + \frac 1c \geq \frac{3}{AM}= \frac 9{a + b + c}$$ Commented Aug 30, 2016 at 17:59
• @mixedmath: the problem with the contest from 2008 Commented Aug 30, 2016 at 18:36
• which contest 2008? Commented Aug 30, 2016 at 18:38
• ru.calameo.com/read/003596578a055905e4b0c Commented Aug 30, 2016 at 19:11

$\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a}=\sum\limits_{cyc}\frac{1}{a(1+a)}+\sum\limits_{cyc}\frac{1}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}\geq\sum\limits_{cyc}\frac{2}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}$.
$(1+abc)\sum\limits_{cyc}\frac{1}{b(1+a)}=\sum\limits_{cyc}\frac{1+abc+b+ab}{b(1+a)}-3=\sum\limits_{cyc}\left(\frac{1+b}{b(1+a)}+\frac{a(1+c)}{1+a}\right)-3\geq$
$\geq2\sum\limits_{cyc}\sqrt{\frac{a(1+b)(1+c)}{b(1+a)^2}}-3\geq6-3=3$.
By the same way $(1+abc)\sum\limits_{cyc}\frac{1}{c(1+a)}\geq3$ and we are done!