If $a+b=c$, then find the maximum value of $d$ when $\frac1a+\frac1b=d$, where $a$, $b$, $c$ are positive integers.
I've tried using basic calculus to find the maximum value of $d$ expressed as a function of $a$.
That is, after some rearrangement, I get $$d=\frac{c}{ab}$$
which can be expressed as the following function (unless, I'm mistaken) $$d(a)=\frac{c}{a(c-a)}$$
where a, b, c $\in \mathbb{R}^+$.
Now, differentiating this function, then solving $d'(a)=0$ should give the value of $a$ at which $d(a)$ takes its maximum value (assuming the maximum and minimum values of $a$ can be ignored in this calculation).
The problem is I get the answer $d=4/c$ which is wrong, since for example, $1/1+1/2\neq4/3$.
Our claim is $1+\frac{1}{c-1}$ is the maximum possible attainable value which attains at $a=1$ and $b=c-1$ or vice versa.
Say $1+\frac{1}{c-1}$ is not the maximum and say the maximum is attained at $(a,b)$ such that $a$,$b$ is greater than $1$ as if one of them equals $1$ then we will get the same maximum. So $\frac{1}{a}\leq \frac{1}{2}$ and $\frac{1}{b}\leq \frac{1}{2}$.
(As $a,b>1$ and they are positive integers).
Hence $\frac{1}{a}+\frac{1}{b}\leq 1$
Hence $1+\frac{1}{c-1}$ is the maximum possible attainable value as it's greater than $1$!