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 $AM \ge GM$ inequality,$$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge \left(4\sqrt[4]{abc}\right)\left(4\sqrt[4]{\frac{1}{abc}}\right)=16$$and equality holds when $1=a=b=c$.

We know :$a>0 \to a+\frac1a \geq 2$ and now; $$(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\\ 1+1+1+1+a+b+c+\frac1a+\frac1b+\frac1c+\frac ab+\frac ac +\frac ba+\frac bc +\frac ca+\frac cb=\\ 4+(a+\frac 1a)+(b+\frac 1b)+(c+\frac 1c)+(\frac ba+\frac ab)+(\frac ac+\frac ca)+(\frac cb+\frac bc)\geq 4+3(2)+3(2)\\ l.h.s \geq 16$$ only $l.h.s=16$ taht $a=b=c=1$so $$a+b+c=1+1+1=3$$