Equivalent of adding to a denominator?

Question

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 $m$?

Answer 1

I think this should do it:

$$\begin{align}\frac nm&\ge a\\ n&\ge am+a-a\\ n&\ge a(m+1)-a\\ \frac n{m+1}&\ge a-\frac a{m+1}\\ \frac{n+1}{m+1}&\ge a+\frac{1-a}{m+1} \end{align}$$

Answer 2

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, $$ a \leq b\cdot c \iff \frac{a}{b} \leq c \qquad \text{ if }\ b>0 \\ a \leq b\cdot c \iff \frac{a}{b} \geq c \qquad \text{ if }\ b<0 $$