<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] In fact we can even prove a general form of this inequality. $$\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)\leqslant(1+n)^n(*)$$
Where: $x_1, x_2, ..., x_n\gt0$ and $x_1+x_2+...+x_n=1$.
Proof:
Firstly let's consider this matrix
\begin{pmatrix}
a_{11} & a_{12} & ... & a_{1n}\\
a_{21} & a_{22} & ... & a_{2n}\\
. & . & . & .\\
. & . & . & .\\
a_{m1} & a_{m2} & ... & a_{mn}
\end{pmatrix}
And prove the following lemma:
$$A(G_1, G_2, ..., G_m)\leqslant G(A_1, A_2, ..., A_n)(**)$$
where $A(x_1, x_2, ..., x_n)=\frac{x_1+x_2+...+x_n}{n}$,
$G(x_1, x_2, ..., x_n)=\sqrt[n]{x_1x_2...x_n}$,
$A_i$ is an arithmetic mean of i-th column of the matrix, $i={1,2,...,n}$.
And $G_j$ is a geometric mean of j-th row of the matrix, $j={1,2,...,m}$.
If any of $A_i=0$ then $(**)$ is obviously true, so let $A_i \gt 0$ then by
the Cauchy–Schwarz inequality we have:$$\sum_{j=1}^n \frac{a_{ij}}{A_j}\geqslant\frac{nG_i}{G(A_1, A_2, ..., A_n)}$$
from ths we get $$\frac{mA_1}{A_2}+\frac{mA_2}{A_2}+...+\frac{mA_n}{A_n}\geqslant\frac{nmA(G_1, G_2, ..., G_m)}{G(A_1, A_2, ..., A_n)}$$ so finally $$A(G_1, G_2, ..., G_m)\leqslant G(A_1, A_2, ..., A_n)$$ which proves our lemma.
Now let's construct following matrix
\begin{pmatrix}
1 & 1 & ... & 1\\
\frac{1}{x_1} & \frac{1}{x_2} & ... & \frac{1}{x_n}\\
\end{pmatrix}
Now
$G(A_1, A_2, ..., A_n)=G(\frac{1+\frac{1}{x_1}}{2}, \frac{1+\frac{1}{x_2}}{2}, ..., \frac{1+\frac{1}{x_n}}{2})=\frac12\sqrt[n]{(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})}$,
$A(G_1, G_2)=A(1, \sqrt[n]{\frac{1}{x_1x_2...x_n}})=\frac12(1+\frac{1}{\sqrt[n]{x_1x_2...x_n}})$
Now using our lemma we have: $$(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})\geqslant \left(1+\frac{1}{\sqrt[n]{x_1x_2...x_n}}\right)^n$$
Then we only need to show that $\frac{1}{\sqrt[n]{x_1x_2...x_n}}\geqslant n\iff \frac{1}{n}\geqslant \sqrt[n]{x_1x_2...x_n}\iff\frac{x_1+x_2+...+x_n}{n}\geqslant\sqrt[n]{x_1x_2...x_n}$ because of our assumption $\sum_{k=1}^n x_k=1$
And the last inequality is true by the AM-GM.
Thus $(*)$ is proven.
</s>