$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ 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 $n\gt2$
 A: 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 $k\geq 2$.
A: I've wanted to give U a little help with the thinking here.
we have the base case: $1-\frac{1}{\sqrt{2}}<\frac{2}{2^{2}} \longrightarrow -\frac{1}{\sqrt{2}}\lt -\frac{1}{2} \\\\\text{and that's true because}\longrightarrow \sqrt{2}\lt 2$
let there be n such that: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$
now we just need to prove the following: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n+1}} \right)\lt \frac{2}{(n+1)^{2}}$
we know that: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}$
so here when we think about the question we end up in a "problem", how to continue.
We have: $\frac{2}{n^2}(1-\frac{1}{\sqrt{n+1}})\leq\frac{2}{(n+1)^2}$
from here what we need to do is the following thing: move $\frac{1}{\sqrt{n+1}}$ to the other side and then group by similarities. from here you'll get the answer. I hope I helped :)
A: For $k\geq2,$ \begin{align} & \frac{2}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\leq \frac{2}{(k+1)^2}\\
\iff  & 1-\frac{k^2}{(k+1)^2}\leq \frac{1}{\sqrt{k+1}}\\
\iff  & \frac{2k+1}{(k+1)^2}\leq \frac{1}{\sqrt{k+1}}\\
\iff  & (2k+1)^2\leq (k+1)^3\\
\iff  & k(k^2-k-1)\geq0\\
\iff  & k^2-k-1\geq0\\
\iff  & k\in\left(-\infty,\frac12-\frac{\sqrt5}2\right]\cup\left[\frac12-\frac{\sqrt5}2,\infty\right)\\
\iff  & k\geq2
\end{align}
Therefore, $$ k\geq2\implies \frac{2}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\leq \frac{2}{(k+1)^2},$$ as required for your induction step.
Addendum
OP:

Hence, we are required to prove
$$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\lt\frac{2}{(k+1)^2},$$ that is, $LHS-RHS<0.$
\begin{aligned}&LHS-RHS\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{1}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\Big(1+\frac{1}{\sqrt{k+1}}\Big)-\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\\
=&\cdots\\ =&\frac{1}{k}-\frac{1}{\sqrt{k+1}}\newline \lt&0,\end{aligned} as required.


@ryang I thought multiplying the expression
by $\left(1-\frac{1}{k+1}\right)$ would change the expression
entirely so it wouldn't be 'equal'?

Yes, so the above attempt is erroneous.

But, writing "$⟺$" with "$<0$", isn't that the
conclusion that I'm trying to reach, and I would think that I can't > just assume that the expression is $<0\,?$

I explained here that to prove an inequality $P(x),$ it is valid and legitimate to rewrite it as a simpler but equivalent inequality $Q(x)$ then explain how $Q(x)$ is true. In effect, the flow of reasoning is this: since $P(x){\iff}Q(x)$ and $Q(x)$ is true, thus $P(x)$ must be true.

Is there potentially a better way to write this step?

All the $<$ ought to be changed to to $\leq$ since the requirement is to prove merely the latter.
Suggestion 1 (this presentation parallels my answer above):
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
Now, for $k\geq2,$
\begin{aligned}&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}\\
\iff &
\frac{1}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\Big(1+\frac{1}{\sqrt{k+1}}\Big)-\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\leq0\\
\iff &\cdots\\ \iff &\frac{1}{k}-\frac{1}{\sqrt{k+1}}\leq0\\ \iff &
k\geq\sqrt{k+1}\\\text{and}\quad &k\geq\sqrt{k+1}.\end{aligned}  Hence, $$k\geq2\implies
\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2},$$ as required.
Suggestion 2 (this presentation corrects your attempt):
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})} \\ \leq&0,\end{aligned} as required.
