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?
Best Answer
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 $\frac56$.