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.

There are lots of ways to prove this.
One way is to square both sides and
cross-multiply.
Another way is to write
$\dfrac{2n+1}{2n+2}
=1-\dfrac{1}{2n+2}
$
and
$\dfrac{\sqrt{n+1}}{\sqrt{n+2}}
=\sqrt{\dfrac{n+1}{n+2}}
=\sqrt{1-\dfrac{1}{n+2}}
$.
Squaring both sides,
you want
$1-\dfrac{1}{n+2}
\ge (1-\dfrac{1}{2n+2})^2
=1-\dfrac1{n+1}+\dfrac1{(2n+2)^2}
$
or
$\dfrac1{(2n+2)^2}
\le \dfrac1{n+1}-\dfrac1{n+2}
=\dfrac1{(n+1)(n+2)}
$
and this is true
for all $n \ge 0$.
Another way is write it as
$\dfrac{2n+2}{2n+1} 
\ge \dfrac{\sqrt{n+2}}{\sqrt{n+1}}
$
or
$1+\dfrac{1}{2n+1} 
\ge \sqrt{1+\dfrac{1}{n+1}}
$.
Squaring,
this is
$1+\dfrac{2}{2n+1}+\dfrac{1}{(2n+1)^2} 
\ge 1+\dfrac{1}{n+1}
$
or
$\begin{array}\\
0
&\le \dfrac{2}{2n+1}-\dfrac{1}{n+1}+\dfrac{1}{(2n+1)^2}\\
&= \dfrac{2(n+1)-(2n+1)}{(2n+1)(n+1)}+\dfrac{1}{(2n+1)^2}\\
&= \dfrac{1}{(2n+1)(n+1)}+\dfrac{1}{(2n+1)^2}\\
\end{array}
$
which is always true.
