Strengthening an inequality I'm reading a book and there's an example problem that goes like this:
Prove that
$$
\left(\frac{1}{2}\right) \left(\frac{3}{4}\right) ... \left(\frac{2n-1}{2n}\right) \le \left(\frac{1}{\sqrt {3n}}\right).
$$
At first the book tries to solve it right away using induction, but gets a false inequality. Then it says that the original inequality is too weak and that we need to strengthen it. The book strengthens the inequality by replacing 3n with 3n+1 so that the inequality turns to 
$$
\left(\frac{1}{2}\right) \left(\frac{3}{4}\right) ... \left(\frac{2n-1}{2n}\right) \le \left(\frac{1}{\sqrt {3n+1}}\right).
$$
Then, by using induction, you get a solid proof. 
Why is this correct? I mean, why can you strengthen an inequality like that? If you had
$$
x \lt 2
$$
and you strengthen the inequality so that it's
$$
x \lt 3,
$$
clearly you can't prove that x is smaller than 2 by proving that it's smaller than 3 (e.g. if x=2,5).
Or is this something that works only when using induction? Or am I just completely misunderstanding this?
 A: Maybe I got it wrong, but I think that the reason for proving
$$ \prod_{k=1}^{n}\frac{2k-1}{2k}\leq \frac{1}{\sqrt{3n}} $$
by proving a stronger inequality, is just that the stronger inequality
$$ \prod_{k=1}^{n}\frac{2k-1}{2k}\leq \frac{1}{\sqrt{3n+1}} $$
is way easier to prove by induction. So it is just a technical reason. 
Anyway, we may also notice that, for any $n\geq 2$:
$$\begin{align*} \left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)^2 &= \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\\&\leq \frac{1}{4n}\prod_{k=2}^{n}\exp\left(\frac{1}{4k(k-1)}\right)\\&\leq \frac{1}{4n}\cdot\exp\left(\sum_{k\geq 2}\frac{1}{4k(k-1)}\right)=\frac{e^{1/4}}{4n}\end{align*} $$
hence:

$$ \forall n\geq 2,\qquad \prod_{k=1}^{n}\frac{2k-1}{2k}\leq \sqrt{\frac{1}{(4e^{-1/4})\,n}}\leq \sqrt{\frac{1}{\left(3+\frac{1}{9}\right) n}}$$

that is even stronger. You may also replace that $3+\frac{1}{9}$ with a $\pi$, but such a result requires some form of the Wallis product to be proved.
