Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$ 
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$

Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?

Probably you mean $a,b,c\geq 0$. You want to find the miminmum of 
$$\ln (1+\frac{1}{a})+\ln(1+\frac{1}{b})+\ln(1+\frac{1}{1-a-b})$$
for $a,b\geq 0$. The partial derivatives are zero iff
$$
-\frac{1}{a(a+1)}+\frac{1}{(1-a-b)(2-a-b)}=0
$$
and a similar equation in which $a$ and $b$ will be replaced. By subtracting, you get $a(a+1)=b(b+1)$ which implies (given that all is positive) $a=b$. But that's the minimum, $a=b=c=1/3$.
