How to approach and solve this proof?

Question

Let e, f, g, h be positive integers.

Let $\frac{e}{f}$ < $\frac{g}{h}$

Show that $\frac{e}{f}$ < $\frac{e+g}{f+h}$ < $\frac{g}{h}$ is possible.

I know that you can say EH < GF, but I don't know what else to do from here. If someone can show me step by step what to do, I think it'd be great for my understanding.

Answer 1

From $\frac{e}f<\frac{g}h$ we have $eh<fg$.

1. $\frac{e}f<\frac{e+g}{f+h}\iff ef+eh<ef+fg\iff eh<fg$
2. $\frac{e+g}{f+h}<\frac{g}h\iff eh+gh<fg+gh\iff eh<fg$

Here cross multiplication is valid since $e, f, g, h$ are positive.

Answer 2

As mentioned, we should use the cross multiplication approach.

$$\frac{e}{f}<\frac{g}{h}$$

Therfore $$eh<gf \tag{1}$$

Now lets deal with the first part of the inequality, namely \begin{align*} \frac{e}{f}&<\frac{e+g}{f+h}\\ e(f+h)&<f(e+g)\\ ef+eh&<fe+fg \tag{substracting $fe$}\\ eh&<fg . \end{align*} Which is actually true for all positive integers $(1)$. Doing the same process with the second part

\begin{align*} \frac{e+g}{f+h}&<\frac{g}{h}\\ (e+g)h&<g(f+h)\\ he+hg&<g f+gh\tag{substracting $gh$}\\he&<gf \end{align*} Which is again true for all positive integers. Hence the statement holds for all cases

$\square$