Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If  $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.
Please help.

By applying Caushy Schwartz inequality we get
( 1 + a + b + c ) ( 1 + 1/a + 1/b + 1/c ) >= 16
Equality holds when a^2 = b^2 = c^2 = 1
Therefore
a = b = c = 1
