$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
alternatively,
$$\prod_{a=2}^{n} {\left(1-\frac{1}{\sqrt{a}}\right)} \lt \frac{2}{n^2} \text{ for } n\geq2$$
I tried to do it by induction but ended up with $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})$ in the RHS which I have no idea what to do with. I also tried to do it with AM/GM inequality but also had little luck. The question did not specifically specify what type of proof to use so I'm assuming you can use any type of proof. The first that came to mind is induction obviously, but I'm really lost on this one. Any help or hints would be appreciated!
With induction, the thing that I'll need to prove is:
$$\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}}) \lt \frac{2}{(1+k)^{2}}$$
But I'm not sure how to approach it.
SOLVED
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
Prove by induction:
A. When $n=2$,
$$\begin{aligned}LHS &= 1-\frac{1}{\sqrt{2}}\newline
&\approx 0.293\newline
RHS&=\frac{2}{2^2}\newline
&=0.5\newline
&\gt LHS\newline
\therefore \text{true for }n&=2\end{aligned}$$
B. Assume the statement is true for the positive integer $n=k$, where $k\in\mathbb{N}$.
That is, assume that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\lt \frac{2}{k^{2}}$$
Now prove the statement for $n=k+1$. That is, prove that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{(k+1)^{2}}$$
By the induction hypothesis:
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{k^{2}}\left( 1-\frac{1}{\sqrt{k+1}} \right)$$
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
For $k\geq2,$ \begin{aligned}&\text{LHS-RHS}\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\left(1+\frac{1}{\sqrt{k+1}}\right)-\frac{1}{(k+1)^2}\left(1+\frac{1}{\sqrt{k+1}}\right)\right)\\
=&\cdots\\ =&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k}-\frac{1}{\sqrt{k+1}}\right)\\
=&\frac{2(\sqrt{k+1}-k)}{k(1+\sqrt{k+1})} \\ \lt&0,\end{aligned} as required.
$\therefore$ It follows from parts A and B by mathematical induction that the result is true for all integers $n\gt2$

Hint: The induction step is just $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})\leq\frac{2}{(k+1)^2}$, which is equivalent to
$$
\frac{1}{k^2}\frac{k}{k+1}\leq\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\iff\frac{k+1}{k}=1+\frac{1}{k}\leq 1+\frac{1}{\sqrt{k+1}}
$$
which boils down to $k\geq\sqrt{k+1}$, for integers $k\geq 2$.
