Need help to prove the last inequality of this mathematical induction Here is the original question:

Let $$x_n=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n - 1}{2n}$$
  Then show that $$x_n\leq\frac{1}{\sqrt{3n+1}} \text{, for all } n = 1, 2, 3, \ldots \text{ .}$$

It is trivial to show that the statement is true for $n=1$.
I assumed it is true for some arbitrary positive integer $k$. Hence, we have
$$x_k\leq\frac{1}{\sqrt{3k+1}}$$
Now, after proper grouping, we arrive at the expression for $x_{k+1}$:
$$
x_{k+1}=x_k\cdot\frac{2k+1}{2k+2}
$$
which implies that
$$
x_{k+1}\leq \frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}
$$
From here, ho do I arrive at the fact that
$$
x_{k+1}\leq \frac{1}{\sqrt{3(k+1)+1}}=\frac{1}{\sqrt{3k+4}}?
$$

Say this inequality is correct for $x_n$. So we can say that: $x_n \leq \dfrac{1}{\sqrt{3n+1}}$ $x_{n+1} = x_n * \dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2}$ So know if we prove the statement below, the inequality is proved: $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}}$ In order to do that, we assume that this statement is correct and then use it to derive an obvious statement. Then we can claim that the initial statement that we had, is correct. $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}} => \dfrac{2n+1}{2n+2} \leq \sqrt{\dfrac{3n+1}{3n+4}} => \dfrac{4n^2+4n+1}{4n^2+8n+4}\leq\dfrac{3n+1}{3n+4}=>$
$12n^3+28n^2+19n+4\leq12n^3+28n^2+20n+4=>n\geq0$
Which is an obvious statement which means for any $n\geq 0$ this statement is correct. So the inequality is proved.
