If positive $a$, $b$, $c$, $d$ satisfy ${1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}=1$, then $abcd\geq 81$

Question

Let $a,b,c,d>0$ satisfying $${1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}=1$$

Prove that $abcd\geq 81$

I've tried to apply arithmetic geometric mean inequality or Cauchy-Schwartz inequality.

$${{1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}\over 4}\geq \left({1\over(a+1)(b+1)(c+1)(d+1)}\right)^{1\over4}$$

or

$$(a+1+b+1+c+1+d+1)\left({1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}\right)\geq 4^2$$

But this wasn't enough to prove $abcd\geq 81$.

Does anyone have ideas??

Answer 1

Because by AM-GM $$\frac{a}{a+1}=\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}\geq\frac{3}{\sqrt[3]{(b+1)(c+1)(d+1)}}.$$ Id est, $$\prod_{cyc}\frac{a}{a+1}\geq\prod_{cyc}\frac{3}{\sqrt[3]{(b+1)(c+1)(d+1)}}$$ or $$\frac{abcd}{\prod\limits_{cyc}(a+1)}\geq\frac{81}{\prod\limits_{cyc}(a+1)}$$ or $$abcd\geq81.$$