Equivalent of adding to a denominator? Given the inequality $\frac{n}{m} \ge \frac{1}{2}$, I want to add $1$ to both $n$ and $m$:
$$\frac{n+1}{m+1}.$$
What would be the equivalent operation on the RHS of the equation?
Adding $1$ to $n$ is equivalent to $+\frac{1}{m}$ on both sides, but what about adding $1$ to $m$?

Hint 1
$$\frac{n}{m}=\frac{n+1}{m+1}+\left(\frac{n}{m}-\frac{n+1}{m+1}\right)$$
and
$$ a \leq b+c \iff a-b \leq c $$
Hint 2
$$\frac{n}{m}=\frac{n+1}{m+1}\cdot\left(\frac{n}{m}\cdot\frac{m+1}{n+1}\right)$$
and,
$$ a \leq b\cdot c \iff \frac{a}{b} \leq c \qquad \text{ if }\ b>0 \\ a \leq b\cdot c \iff \frac{a}{b} \geq c \qquad \text{ if }\ b<0 $$
