<s>[INST] 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?
 [/INST] Here is a "simple" proof utilizing GM-HM inequality.
Use GM-HM inequality on the set $\{\frac{a+1}{a},\frac{b+1}{b},\frac{c+1}{c}\}  $ to get:
$$\left(\,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\,\right)^\frac{1}{3} \geq \frac {3} {\left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)} $$
$$\implies \,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right) \geq \frac {27} {\left( \,\, \left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)\,\,\right)^{3}} \,\,\,(♦)$$
Now, maximizing $\left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)$ according to the constraint $a+b+c=1$ , its maximum value comes out to be  $\frac{3}{4}$.
$$\therefore \left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right) \leq \frac {3}{4}\,\,\,(♣)$$
Plug $(♣)$ in $(♦)$ to get : 
$$\left(\frac {a+1}{a}\right) \left( \frac{b+1}{b}\right) \left(\frac{c+1}{c}\right) \geq \frac {27} {(3/4)^3} =\frac {27\times 64}{27}  = 64$$
$$OR$$
$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \geq 64$$
as desired. 
$$HENCE \,\,\,\, PROVED$$
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