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

Question

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

Answer 1

By Rearrangement

$\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a}=\sum\limits_{cyc}\frac{1}{a(1+a)}+\sum\limits_{cyc}\frac{1}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}\geq\sum\limits_{cyc}\frac{2}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}$.

Now by AM-GM

$(1+abc)\sum\limits_{cyc}\frac{1}{b(1+a)}=\sum\limits_{cyc}\frac{1+abc+b+ab}{b(1+a)}-3=\sum\limits_{cyc}\left(\frac{1+b}{b(1+a)}+\frac{a(1+c)}{1+a}\right)-3\geq$

$\geq2\sum\limits_{cyc}\sqrt{\frac{a(1+b)(1+c)}{b(1+a)^2}}-3\geq6-3=3$.

By the same way $(1+abc)\sum\limits_{cyc}\frac{1}{c(1+a)}\geq3$ and we are done!