2020/020) Given a,b,c are positive and $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} = 2$ show that $(abc) <= \frac{1}{8}$

we have $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} = 2$
Hence
$(1+b)(1+c) + (1+c)(1+a) + (1+a)(1+b) = 2(1+a)(1+b)(1+c)$
or $1 + b + c + bc + c + a + ac + 1+ a + b + ab = 2 + 2a + 2b + 2c + 2ab + 2bc + 2ac + 2abc$
or $2abc + ab + bc + ca = 1$
Applying AM GM inequality we get
$\frac{1}{4}(ab+bc+ca+2abc) >= \sqrt[4]{2abc.ab.bc.ca}$
Or $\frac{1}{4} >= \sqrt[4] {2 a^3b^3c^3}$
Or $\frac{1}{256} >= 2(abc)^3$
or $(abc) <= \sqrt[3]\frac{1}{512}$
or  $(abc) <= \frac{1}{8}$