Prove by induction $\left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$ Prove by induction $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$
What would be the best way to solve this by the induction method?
 A: Hint: Prove that
$$
\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)
\left(1-\frac{1}{2^{n+1}}\right)
\ge
\frac{1}{4}+\frac{1}{2^{n+2}}
$$
A: To prove a statement by induction, we must first prove the statement works for $n=1$, then extend that statement by supposing that the given statement is true for $n$, then proving it maintains truth for $n+1$. I am also assuming $n$ is an positive integer.
The logic behind that is if the statement works for $n=1$, then it will work for $n+1=2$, and if it works for $n=2$... well you get the point.
In this case for $n=1$, we have $(1- \frac{1}{2})\geq \frac{1}{4} + \frac{1}{2^{(1)+1}}$
or $\frac{1}{2} \geq \frac{1}{4} + \frac{1}{4}$ which is true.
Now since that statement is true for $n=1$, its time to do the induction step.
We can now assume $(1- \frac{1}{2^n})\geq \frac{1}{4} + \frac{1}{2^{n+1}}$ holds true for n.
Consider $n+1$:
$(1- \frac{1}{2^{n+1}})\geq \frac{1}{4} + \frac{1}{2^{(n+1)+1}}$
Simplifying gives you $(1- \frac{1}{2^{n+1}})\geq \frac{1}{4} + \frac{1}{2^{n+2}}$
Taking it one step further: $1- \frac{1}{2}\cdot\frac{1}{2^{n}}\geq \frac{1}{4} + \frac{1}{4}\cdot\frac{1}{2^{n}}$
$1\geq \frac{1}{2^n}$ or $2^n \geq 1$, which is true because n is a positive integer.
So if $(1- \frac{1}{2^n})\geq \frac{1}{4} + \frac{1}{2^{n+1}}$ is true for $n$, it will work for $n+1$
And that's all you need to prove an induction.
A: $$
(1-\delta)(1-\varepsilon) = 1 - \delta - \varepsilon + \delta\varepsilon > 1 - (\delta + \varepsilon)
$$
