What is the biggest possible value of $\frac 1a+\frac 1b$ given that $2 <\frac ab < 7$ where $a$ and $b$ are distinct integers and $a + b = 10$. Let $a$ and $b$ be distinct integers that satisfy $2 < \frac ab < 7$. If $a + b = 10$, then what is the biggest possible value of $\frac 1a +\frac 1b$?
It's easy to show that $\frac 1a +\frac 1b=\frac {10}{ab}$ for some $a$ and $b$. Now we just need to find $ab$. However, without finding the interval solutions, how do we get $ab$ from $2 < \frac ab < 7$? Or maybe we don't need to find the value of $ab$? This is where I get stuck and I appreciate any help.
 A: HINT: Suggested method for solving:
First, notice that $\frac ab$ is well defined, so $b$ is nonzero.  It is positive, so $a$ is nonzero.
Second, figure out which of these can be true: 1) $a$ and $b$ are both positive 2) $a$ is positive and $b$ is negative 3) $a$ is negative and $b$ is positive 4) $a$ and $b$ are both negative
Third, for each one that can be true, substitute $b=10-a$ into the inequality, and split the inequality into two pieces: one for $2<\frac ab$ and one for $\frac ab<7$
Fourth, solve each of these pairs of inequalities, to get possible domains for $a$.
Fifth, if there aren't many possible values for $a$, go through each one to find the maximum of $\frac 1a+\frac 1b$.  If there are a lot of possible values for $a$, use the derivative of $\frac 1a+\frac 1b$, and check only values adjacent to extreme points or to the edges of each of the domain.
A: If you assume that $a$ and $b$ are positive, then there are only nine possible combinations with $a + b = 10$, and of these, only two also meet $2 < \frac{a}{b} < 7$:

*

*$a = 7, b = 3 \implies \frac{1}{a} + \frac{1}{b} = \frac{10}{21} \approx 0.47619$

*$a = 8, b = 2 \implies \frac{1}{a} + \frac{1}{b} = \frac{5}{8} = 0.625$
Clearly, the latter is bigger.

But let's consider the possibility of negative numbers.  We're given that $a + b = 10$, so let's just make the substitution $b = 10 - a$ everywhere so that there's only one variable to work with.  Then your question becomes:

What is the biggest possible value of $\frac{1}{a} + \frac{1}{10-a}$ given that $2 < \frac{a}{10-a} < 7$?

Obviously $a \ne 10$.
If $a < 10$, then $10 - a$ is positive, so we can multiply the inequality by it without changing the order.
$$2(10 - a) < a < 7(10 - a)$$
$$20 - 2a < a \text{ and } a < 70 - 7a$$
$$20 < 3a \text{ and } 8a < 70$$
$$a > \frac{20}{3} = 6.666... \text{ and } a < \frac{70}{8} = 8.75$$
The only integer values of $a$ meeting this inequality are 7 and 8, which I've already worked out above.
Alternatively, if $a > 10$, then $10 - a$ is negative, so multiplying by it flips the order.
$$2(10 - a) > a > 7(10-a)$$
$$20 - 2a > a \text{ and } a > 70 - 7a$$
$$20 > 3a \text{ and } 8a > 70$$
$$a < \frac{20}{3} = 6.666... \text{ and } a > \frac{70}{8} = 8.75$$
But this is a contradiction, so this case adds no new possible values for $a$ (or $b$).  So we still have $a \in \{ 7, 8 \}$.  And as already shown, $a = 8$ maximizes the target expression.
