For a given condition is it true that $a+b+c=3$. Suppose a, b, c  are positive real numbers such that  
$$(1+a+b+c)\left(1+\frac 1a+\frac 1b+\frac 1c\right)=16$$ 
Then is it true that  we must have  $a+b+c=3$ ?
Please help me to solve this. Thanks in advance. 

By Cauchy-Schwarz $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq\left(1\cdot1+\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}}\right)^2= 16$$
The equality occurs for $a=b=c=1$, which says that $a+b+c=3$.
