<s>[INST] Proving by induction inequalities that lack the variable on the right side: $\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$ Doing proof by induction exercises with inequalities, I got stuck on one that is a bit different from the others. There is no $n$ term on the rightmost part of the inequality:

Prove that the following holds for all $n \ge 1$:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$

All the other proofs I did before had some $n$ involved there, but now there is none. I wonder how will this difference affect my standard attempt:

Test for $n = 1$:
$$\frac{1}{3} \le \frac{5}{6} \implies 6 \le 15$$

We have to prove that it holds for $n + 1$, that is:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$

Asssume
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$

We start off with
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
And now I see the problem. Normally, I apply the hypothesis first on this part:
$$\color{blue}{\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
But all it yields is this:
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} + \frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
If I move $\frac{1}{(n + 1)+(n+2)}$ to the other side, all I get is
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} \le \frac{5}{6} - \frac{1}{(n + 1)+(n+2)}$$
Which cannot be guaranteed.
How can I prove this, then?
 [/INST] we assume 
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
looking for $n+1$ case, we see :
$$\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n+1)+(n+2)}=\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}-\frac{1}{n+2}\leq \frac{5}{6}-\frac{1}{n+2}\leq \frac{5}{6}$$
</s>