An algebra problem about minimum value

$(a+1)(b+1)(c+1)\left(\dfrac 1a + 1 \right)\left(\dfrac 1b + 1 \right)\left(\dfrac 1c + 1 \right) \\ = (a+1)(b+1)(c+1)\left(\dfrac {1+a}a \right)\left(\dfrac {1+b}b \right)\left(\dfrac {1+c}c \right) \\ = \left(\dfrac {(1+a)^2}a \right)\left(\dfrac {(1+b)^2}b \right)\left(\dfrac {(1+c)^2}c \right) \\ = \left(\color{#3D99F6}{\dfrac 1a} + 2 + \color{#3D99F6}{a} \right)\left(\color{#3D99F6}{\dfrac 1b} + 2 + \color{#3D99F6}{b} \right)\left(\color{#3D99F6}{\dfrac 1c} + 2 + \color{#3D99F6}{c} \right) \quad \quad \small \color{#3D99F6}{\text{By AM-GM inequality: }\dfrac 1a + a \ge 2} \\ \color{#3D99F6}{\ge} \left(\color{#3D99F6}{2} + 2 \right) \left(\color{#3D99F6}{2} + 2 \right) \left(\color{#3D99F6}{2} + 2 \right) = \boxed{64}$

Use Holder's inequality...

$\large (a+1)^{\frac{1}{6}}(b+1)^{\frac{1}{6}}(c+1)^{\frac{1}{6}}(\frac{1}{a}+1)^{\frac{1}{6}}(\frac{1}{b}+1)^{\frac{1}{6}}(\frac{1}{c}+1)^{\frac{1}{6}} \geq \sqrt[6]{\frac{abc}{abc}} + \sqrt[6]{1}$

Giving us $\large \sqrt[6]{(a+1)(b+1)(c+1)(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1)} \geq 2$

Raising both sides to the 6th power gives us

$(a+1)(b+1)(c+1)(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \geq \boxed{64}$ .

We want a+1,b+1&c+1 to be minimum. Since a,b and c are positive, we have a=b=c=1, giving the value 64.