How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$? So this was part of a bigger induction proof problem, but I'm stuck at what I think is the last step. I need to show that
$\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$, for $n \ge 1$
Would it be enough to show that the limit as it approaches positive infinity is both 1, and that since the inequality is true for $n=1$, and $1/2 < \sqrt{1/2}$, the function on the right is always greater? I'm pretty sure this isn't the way to go about it but I'm really stuck.

Although this is a little indirect, it keeps calculation to a
minimum.
For any positive integer $m,$ we have
$$
(m - 1)(m + 1) = m^2 - 1 < m^2,
$$
therefore
$$
\frac{m - 1}{m^2} < \frac1{m + 1},
$$
therefore
$$
\left(\frac{m - 1}m\right)^2 = \frac{(m - 1)^2}{m^2} \leqslant
\frac{m - 1}{m + 1} = 1 - \frac2{m + 1} < 1 - \frac2{m + 2} =
\frac{m}{m + 2},
$$
therefore
$$
\frac{m - 1}m < \sqrt{\frac{m}{m + 2}}.
$$
Let $n$ be a non-negative integer.
Taking $m = 2n + 2,$ and simplifying, we get
$$
\frac{2n + 1}{2n + 2} < \sqrt{\frac{2n + 2}{2n + 4}}
= \sqrt{\frac{n + 1}{n + 2}} = \frac{\sqrt{n + 1}}{\sqrt{n + 2}}.
$$
