If $x_1, \ldots, x_6$ are positive real numbers that add up to $2$. Show that: 
If $x_1,x_2,x_3,x_4,x_5$ and $x_6$ are positive real numbers that add up to $2$, then:
$$2^{12} \leq \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right) .$$

Factoring out the $\dfrac{1}{x_1}, \ldots, \dfrac{1}{x_6}$ from each term, I get $$\dfrac{1}{x_1}(x_1 + 1)\dfrac{1}{x_2}(x_2 + 1)\dfrac{1}{x_3}(x_3 + 1)\dfrac{1}{x_4}(x_4 + 1)\dfrac{1}{x_5}(x_5 + 1)\dfrac{1}{x_6}(x_6 + 1)$$
We know that: $x_1 + \cdots + x_6 = 2$
Also, I see that $x_1, \ldots, x_6$ are small numbers and $1$ over something small will give something big.

How can I keep going?

 A: The inequality is equivalent to
$$ \frac{1}{ \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right)} \leq \frac{1}{2^{12}}$$
or
$$\sqrt[6]{\prod \frac{x_i}{x_i+1}} \leq \frac{1}{4}$$
Now, by AM-GM we have
$$\sqrt[6]{\prod \frac{x_i}{x_i+1}}  \leq \frac{\sum \frac{x_i}{x_i+1}} {6}$$, so it suffices to prove
$$\sum \frac{x_i}{x_i+1} \leq \frac{3}{2}$$
This is equivalent to
$$6- \sum \frac{1}{x_i+1} \leq \frac{3}{2} \Leftrightarrow \sum \frac{1}{x_i+1} \geq \frac{9}{2}$$
But this is Just Cuachy Schwartz
$$6^2 \leq \left( \sum \frac{1}{x_i+1} \right) \left(\sum (x_i+1)\right)=\left( \sum \frac{1}{x_i+1} \right) \left(2+6\right)$$
A: The function
$$f(x):=\log\left(1+{1\over x}\right)\qquad(x>0)$$
has a positive second derivative, whence it is convex. Jensen's inequality then implies
$$\sum_{k=1}^6{1\over 6}\log\left(1+{1\over x_k}\right)\geq\log\left(1+{1\over\sum_{k=1}^6{1\over6} x_k}\right)\ .\tag{1}$$
As $\sum_{k=1}^6 x_k=2$ the right side of $(1)$ evaluates to $\log 4$, so that we immediately obtain
$$\prod_{k=1}^6\left(1+{1\over x_k}\right)\geq 4^6=2^{12}\ .$$
A: If
$$\sum^{n}_{i=1}x_i=S$$
By $AM\ge GM$, we have
$$1+x_i=\frac{Sx_i+\sum^{n}_{k=1}x_k}{S} \ge\frac{n+S}{S}\sqrt[n+S]{x_i^s\prod^{n}_{k=1}x_k}$$
$$\implies \frac{1+x_i}{\sqrt[n+S]{x_i^S\prod^{n}_{k=1}x_k}}\ge \frac{n+S}{S}$$
$$\implies \prod^{n}_{i=1}{\frac{1+x_i}{\sqrt[n+S]{x_i^S\prod^{n}_{k=1}x_k}}}\ge \left(\frac{n+S}{S}\right)^n$$
$$\implies \prod^{n}_{i=1}\frac{1+x_i}{x_i}\ge \left(\frac{n+S}{S}\right)^n$$
Now plug in $n=6,S=2$ and be forever happy.
