Mathematical induction for inequalities with a constant at the right side Thanks to How to use mathematical induction with inequalities? I can now work with some induction problems related to inequalities. I've been following the logic that I was presented there so far.
Until for $n\ge1$, I had to prove the following:
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+(n+1)}\le\frac{5}{6}$$
Unlike the last time, now I have a constant $\frac{5}{6}$ at the right side. Maybe it makes no difference, so despite that I decided to try it anyway with pretty much the same procedure that André Nicolas used in How to use mathematical induction with inequalities?:
The inequality holds for $n=1$. We now assume the following:
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+(n+1)}\le\frac{5}{6}$$
We want to prove that it holds for $n+1$, so we'd like to show this:
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+(n+1)}+\frac{1}{(n+1)+(n+2)}\le\frac{5}{6}$$
(So something different happened since last time: the right side did not change at all. It worried me, but since I'm just following the same steps I always do I will leave it like that and see what happens).
By the induction assumption we can say that
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+(n+1)}+\frac{1}{(n+1)+(n+2)}\le\frac{5}{6}+\frac{1}{(n+1)+(n+2)}$$
We will be finished if we can show that
$$\frac{5}{6}+\frac{1}{(n+1)+(n+2)}\le\frac{5}{6}$$
.... Oh wait. That's impossible! (I guess)

I guess that it is because the right side remained unchanged back then. But well, in that case, I'm not sure how should I do it then. My question then: how would I solve mathematical induction problems with inequalities that have a constant in the right side?
 A: You actually want to show 
$$\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{(n+1)+(n+1)}+\frac{1}{(n+1)+(n+2)}\le\frac{5}{6}$$
So you take the inductive hypothesis, and subtract $\frac{1}{n+1}$ from and add $\frac{1}{(n+1)+(n+1)}+\frac{1}{(n+1)+(n+2)}$ to the left hand side.  Since you can show that change is less than zero, the overall sum is still less than or equal to $\frac56$.
A: When you say "What we want to prove is ...", you're actually wrong, you have to "update" the $n$ to $n+1$ on every fraction, not only the one you add.
A: Hint $\ \ $ If  $\rm\ \ f(n) =\, \dfrac{5}6 - \dfrac{1}{n\!+\!1} - \dfrac{1}{n\!+\!2} - \,\cdots\, - \dfrac{1}{2n\!+\!1}\ $ then $$\rm f(n\!+\!1)-f(n) = \dfrac{1}{n\!+\!1}-\dfrac{1}{2n\!+\!2}-\dfrac{1}{2n\!+\!3} = \dfrac{1}{2(n\!+\!1)(2n\!+\!3)} > 0$$
thus $\rm\:f(n)\:$ is increasing, so $\rm\:f(n) \ge \cdots \ge f(1) \ge f(0) = \frac{5}6\!-\!\frac{1}2\!-\!\frac{1}3 = 0\:$ by induction.
Remark $\ $ In the same way, one can often reduce inductive proofs to trivial inductions, such as the above induction that an increasing function stays $\ge$ its initial value.  For further discussion and examples see my many prior posts on telescopy. There, you'll learn that, by summing, the above proof can be viewed as a trivial induction that a sum of nonnegative integers stays nonnegative. The sooner one learns how to restructure  inductive proofs to make the induction "obvious", the sooner one will be able to tackle much more complicated inductions that arise in the wild.
