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?
 A: By AM-GM we have:
$1 + \frac{1}{a} = \frac{1}{a}(a + b + c + a) \ge \frac{1}{a}4\sqrt[4]{{{a^2}bc}}$
$\Rightarrow 1 + \frac{1}{a} \ge \frac{4}{a}\sqrt[4]{{\frac{{{a^4}bc}}{{{a^2}}}}} = 4\sqrt[4]{{\frac{{bc}}{{{a^2}}}}} $
And $1 + \frac{1}{b} \ge 4\sqrt[4]{{\frac{{ca}}{{{b^2}}}}};1 + \frac{1}{c} \ge 4\sqrt[4]{{\frac{{ab}}{{{c^2}}}}}$
$$\Rightarrow \left( {1 + \frac{1}{a}} \right)\left( {1 + \frac{1}{b}} \right)\left( {1 + \frac{1}{c}} \right) \ge 4\sqrt[4]{{\frac{{bc}}{{{a^2}}}}}4\sqrt[4]{{\frac{{ca}}{{{b^2}}}}}4\sqrt[4]{{\frac{{ab}}{{{c^2}}}}} = 64$$
A: Use the AM GM inequality :     $$a+a+b+c\geq 4(a^2bc)^\frac{1}{4}$$ $$a+b+b+c\geq 4(ab^2c)^\frac{1}{4}$$ $$a+b+c+c\geq 4(abc^2)^\frac{1}{4}$$ Multiply the three inequalities and then divide $abc$ on both sides to get the desired inequality.
A: 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. 
$$HENCE \,\,\,\, PROVED$$
A: Sheer brute force is adequate:
$$(2a + b + c)(a + 2b + c)(a + b + 2c)$$
$$= 2(a^3 + b^3 + c^3) + 7(a^2b + ab^2 + a^2c+ac^2 + b^2c+bc^2) + 16(abc)$$
$$\geq 2(3abc) + 7(6abc) + 16(abc) = 64abc$$
A: 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, $a=b=c=1/3$.
A: 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.
A: We need prove: $(1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3$
$\Leftrightarrow 1+abc+ab+bc+ca+a+b+c \geq 1+3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}+abc$
$\Leftrightarrow ab+bc+ca+a+b+c \geq 3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}$
Right by AM-GM. Apply we have:
$\left(1+\frac{1}{a} \right)\left(1+\frac{1}{b} \right)\left(1+\frac{1}{c} \right)=\dfrac{(1+a)(1+b)(1+c)}{abc} \geq \dfrac{(1+\sqrt[3]{abc})^3}{abc} \geq 64$
From $a+b+c=1 \Rightarrow abc\le \frac{1}{27}$
$$\Rightarrow \dfrac{(1+\sqrt[3]{abc})^3}{abc}=\bigg(\dfrac{1}{\sqrt[3]{abc}}+1\bigg)^3 \geq 64$$
When $a=b=c=\dfrac{1}{3}$
