Equality with Inequality

Relevant wiki: Titu's Lemma

$\begin{aligned} (1+a+b+c)\color{#3D99F6}\left(1+\frac 1a+\frac 1b + \frac 1c\right) & \ge (1+a+b+c){\color{#3D99F6}\left(\frac {(1+1+1+1)^2}{1+a+b+c}\right)} = 16 & \small \color{#3D99F6} \text{By Titu's lemma} \end{aligned}$

Equality occurs when $a=b=c=1$ . Therefore, when $(1+a+b+c)\left(1+\dfrac 1a+\dfrac 1b + \dfrac 1c\right) = 16$ , $a+b+c = \boxed{3}$ .

By $\color{#3D99F6}{\text{AM-GM inequality}}$ on $(1+a+b+c)$ we have:

$(1+a+b+c)\ge 4 \sqrt[4]{abc} ~~-~~\text{(1)}$

Now, by $\color{#3D99F6}{\text{AM-GM inequality}}$ on $\left(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$ we have:

$\left(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 4\sqrt[4]{\dfrac{1}{abc}} ~~-~~\text{(2)}$

By $\text{(1)}\times \text{(2)}$ with corresponding sides we get:

$\left(1+a+b+c\right)\left(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 16$

Equality occurs when $a=b=c=1$

Hence, $a+b+c=\boxed{3}$