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.
 A: What follows gives a stronger  result in a natural way. At the end I give the strongest possible form of the inequality.
The inequality in question can be stated in a nicer form (two symbols less) by replacing $n$ with $n-1$
$$ \sqrt{n\over n+1}\ge {2n-1\over 2n},\quad n\ge 1\qquad (*)$$
There is a simple inequality $$\sqrt{1+x}\le 1+{x\over 2},\quad x\ge -1$$ verified  immediately by squaring both sides. Thus $${\sqrt{n+1\over n}=\sqrt{1+{1\over n}} \le 1 +{1\over 2n}={2n+1\over 2n}}$$ Therefore
$$\sqrt{n\over n+1}\ge {2n\over 2n+1}\qquad (**)$$ The inequality $(**)$ is stronger than $(*)$ as $${{2n\over 2n+1}=1-{1\over 2n+1} \ge 1-{1\over 2n}={2n-1\over 2n}}$$
Remark We can look for the strongest form of the inequality $(**),$ namely
$$\sqrt{n\over n+1}\ge {2n+a\over 2n+a+1},\qquad a>0$$ On squaring  both sides and applying cross multiplication we eventually end up with an equivalent affine inequality
$$(1-2a)n\ge a^2$$ Therefore
$$n\ge {a^2\over 1-2a},\quad a<{1\over 2}$$
The inequality holds for $n\ge 1$ if $a\le \sqrt{2}-1.$ Hence
$$\sqrt{n\over n+1}\ge {2n+\sqrt{2}-1\over 2n+\sqrt{2}}\qquad (***)$$ is the strongest possible form.
A: 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.
A: Let $f(x)= \frac{2n +x}{\sqrt{n+x}}$ and, for $x>0$, evaluate
$$f’(x) = \frac x{2(n+x)^{3/2}}\ge 0
$$
So, $f(x)$ is an non-decreasing function of $x$, which gives
$f(1)\le f(2)$ and equivalently
$$\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}
$$
A: 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
$$
\frac{2n + 1}{2n + 2} < \sqrt{\frac{2n + 2}{2n + 4}}
= \sqrt{\frac{n + 1}{n + 2}} = \frac{\sqrt{n + 1}}{\sqrt{n + 2}}.
$$
A: As noticed by @marty cohen in his answer and @Kurt G. in the comments, the more natural and simple way in this case is by algebraic manipulation, since all terms are positive we have
$$
\require{cancel}
\begin{align}
\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}\\\\
&\iff (2n+1)\sqrt{n+2}\le(2n+2)\sqrt{n+1}\\\\
&\iff (2n+1)^2(n+2)\le(2n+2)^2(n+1)\\\\
&\iff \cancel{4n^3}+\cancel{12n^2}+9n+2\le \cancel{4n^3}+\cancel{12n^2}+12n+4\\\\
&\iff 3n+2\ge 0\\
\end{align}$$
A: For $N\geq 1$, both expressions are positive and less than 1. So we may compare their reciprocals; the one with smaller reciprocal is larger. For the first one, $${(\frac{2n+1}{2n+2})}^{-1}=\frac{2n+2}{2n+1}=1+\frac{1}{2n+1};$$ while $${(\frac{\sqrt{n+1}}{\sqrt{n+2}})}^{-1}=\frac{\sqrt{n+2}}{\sqrt{n+1}}=\frac{\sqrt{(n+2)(n+1)}}{n+1} <\frac{n+\frac{3}{2}}{n+1}=1+\frac{1}{2(n+1)} < 1+\frac{1}{2n+1}.$$ The result follows.
