<s>[INST] 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.
 [/INST] 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.
</s>