The Meaning of Life

The smallest possible value for $a$ is $43,$ which gives us the equation

$\dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{42} - \dfrac{1}{43} = \dfrac{1}{1806}$

$\Longrightarrow 1806(b + c) = bc \Longrightarrow (b - 1806)(c - 1806) = 1806^{2}.$

We can then maximize $c$ by letting $b - 1806 = 1$ and

$c - 1806 = 1806^{2} \Longrightarrow c = 1806^{2} + 1806 = 1806*1807 = \boxed{3263442}.$

Constraints on $a,\:b$

By definition we have $1\leq a\leq b\leq c$ . Let's find some additional constraints on $a$ : $\begin{aligned} 0&< \frac{1}{b}+\frac{1}{c}=\frac{1}{42}-\frac{1}{a} &\Rightarrow &&a&\geq 43,&&&&& \frac{1}{42} &=\frac{1}{a} +\frac{1}{b} +\frac{1}{c} \leq \frac{3}{a} &\Rightarrow && a\leq 3\cdot 42=126 \end{aligned}$ We now know $a\in\{43;\:\ldots;\:126\}$ . Lets find a lower bound for $b$ : $\begin{aligned} &&0&<\frac{42ab}{c}=42ab\left(\frac{1}{42}-\frac{1}{a}-\frac{1}{b}\right)=ab-42(a+b)=(a-42)(b-42)-42^2 \\\\ \Rightarrow && 1+42^2 &\leq (a-42)(b-42) \quad\Rightarrow \quad b \geq 42+\frac{1765}{a-42} \end{aligned}$ Together with $b\geq a$ we have two lower bounds for $b$ we may combine into $l(a)$ and get $\begin{aligned} l(a)&:=\max\left\{a;\:42+\frac{1765}{a-42}\right\} &\Rightarrow &&l(a) &\leq b\leq c \end{aligned}$ Let's summarize our results into the set $\Omega$ of remaining values for $a,\:b$ : $\begin{aligned} \Omega&:=\left\{(a,\:b)\in\mathbb{N}^2|\:43\leq a\leq 126, \quad l(a)\leq b\right\} &&&\text{and}&&\frac{1}{c}=\frac{1}{42}-\frac{1}{a}-\frac{1}{b}=:\frac{1}{f(a,\:b)} \end{aligned}$ However, it is really difficult to solve this problem over the discrete set. Instead, we will consider a larger set $\Omega'$ with the same borders but $a,\:b,\:c\in\mathbb{R}$ .

Global Maximum for $c$

We now want to maximize $c=f(a,\:b)$ on $\Omega'$ . For points within $\Omega'$ we notice $\begin{aligned} \bullet &&f(a,\:b) &>f(a,\:b') &\text{for}&& b&<b'\\ \bullet &&f(a,\:b)&>f(a',\:b) &\text{for}&& a&< a' \end{aligned}$ i.e. $f$ strictly increases whenever we move down or to the left within $\Omega'$ . Our maximum (if it exists) must lie on the lower left border $\delta\Omega$ defined by $\begin{aligned} \delta\Omega &:= \left\{ (43,\:b)\in\mathbb{R}^2|\: b\geq l(43) \right\}\cup\left\{ (a,\:l(a))\in\mathbb{R}^2|\:a \in[43;\:84] \right\} \end{aligned}$ The first part of the border is easy to check - $f(43,\:b)$ increases as we move down and our only candidate is $(43,\:l(43))$ . It's also possible to show for the second part that $f(a,\:l(a))$ increases as we move to the left so we get the global maximum of $f$ on $\Omega'$ at $\begin{aligned} a&=43, & b&=l(43)=42+1765=1807 &&&\Rightarrow&&&& c&=\frac{1}{f(a,\:b)}=\boxed{3263442} \end{aligned}$ The solution is integer-valued and also satisfies $b\leq c$ , so it is indeed a solution to the problem.

The smallest possible value for a is 43

$\frac{1}{b}$ + $\frac{1}{c}$ = $\frac{1}{42}$ - $\frac{1}{43}$ = $\frac{1}{42*43}$

The smallest possible value for b will be 42*43+1

$\frac{1}{c}$ = $\frac{1}{(42*43)*(42*43+1)}$

Therefore, the maximum $c =$ $(42*43)*(42*43+1)$ = $\boxed{3263442}$