discrete mathematics - Proof by Mathematical Induction for Inequality

Prove, by mathematical induction, that $$\sum_{i=1}^n\frac{i}{i+1}\leq \frac{n^2}{n+1}$$
(When $n$ is a natural number.)

So I went through all the typical steps you'd go through with mathematical induction, and after assuming that the statement at $n+1$ was true, I ended up with the following:

$$\sum_{i=1}^{n+1}\frac{i}{i+1}\leq \frac{(n+1)^2}{n+2}$$
... (algebra) ...

$$\sum_{i=1}^n\frac{i}{i+1}\leq \frac{n^2+n}{n+2}$$

But since the right hand side of the inequality I ended up with is greater than the original right hand side of the inequality I wanted to end up with, this doesn't help me prove the original statement. So I've either done something wrong with my algebra or I'm not thinking far enough outside the box; how do I use induction to prove this?

Answer

Suppose that $\sum_{i=1}^{i=n}{i\over{i+1}}\leq {n^2\over{n+1}}$,

$\sum_{i=1}^{i=n+1}{i\over{i+1}}\leq {n^2\over{n+1}}+{{n+1}\over{n+2}}$.

${n^2\over{n+1}}+{{n+1}\over{n+2}}=$

${{n^2(n+2)+(n+1)^2}\over{{(n+1)(n+2)}}}$

$\leq {{n(n^2+2n+1)+(n+1)^2}\over{{(n+1)(n+2)}}}$

$={{n(n+1)^2+(n+1)^2}\over{(n+1)(n+2)}}={{(n+1)^2}\over{n+2}}$.