Proving $\frac{n}{n+1} < \frac{n+1}{n+2}$ by induction? I have the inequality
$\frac{n}{n+1} < \frac{n+1}{n+2}$
I'm not sure how to go about proving it. I've started by testing with n = 1, which results in
$\frac{1}{2} < \frac{2}{3}$ which is true
I then assume true for n = k and have to prove that it is true for n = k + 1, but I don't know how to start manipulating
$\frac{k}{k+1} < \frac{k+1}{k+2}$
to become
$\frac{(k+1)}{(k+1)+1} < \frac{(k+1)+1}{(k+1)+2}$
How do I go about doing this?
 A: You don't need induction, since it is enough to show
$$n(n+2)<(n+1)^2$$
A: This only requires basic inequation manipulation ($n \in \mathbb{N}$):
$${n \over n+ 1} < {n + 1\over n+ 2}$$
$$\iff n(n+2) < (n + 1)^2$$
$$\iff n^2+2n < n^2 + 2n + 1$$
$$\iff 0 <  1$$
Done.
A: Note that
$$ \frac{n}{n+1} = 1 - \frac{1}{n+1},$$
$$ \frac{n+1}{n+2} = 1 - \frac{1}{n+2},$$
so actually you just need to show that
$$\frac{1}{n+1} > \frac{1}{n+2}.$$
A: Use : if $ 0 < a < b \Rightarrow \dfrac{a}{b} < \dfrac{a+1}{b+1} $, with $a = k, b = k+1$, and once more $ a = k+1, b = k+2$.
A: It is a well-known inequality that, if $a,b,h>0$ then
$$\begin{cases}
\dfrac ab<1\implies \dfrac ab<\dfrac{a+h}{b+h}<1,\\
\dfrac ab>1\implies \dfrac ab>\dfrac{a+h}{b+h}>1.
\end{cases}$$
A: if you need induction:
\begin{gather*}
n(n+2)<(n+1)^2 \\
\begin{aligned}
(n+1)(n+3)={}&n(n+2+1)+(n+3)=(n(n+2)+n+n+3)<(n+1)^2+2n+3={} \\
{}={}&(n^2+4n+4)=(n+2)^2.
\end{aligned}
\end{gather*}
A: Here is another way: Let $f(x) = { x \over x+1}$ and show that
$f'(x) >0 $ for all $x \ge 0$. Then $f$ is strictly increasing and
so $f(n) < f(n+1)$ for all $n \ge 0$.
