Let $m$, $n$, and $p$ be real numbers such that $n\gt m\gt 0$ and $p\gt 0$. Prove that
$$\frac{m+p}{n+p}\gt\frac{m}{n}$$
My attempt:
$$\frac{m}{n}+\frac{n+p}{n+p}\gt\frac{m}{n}\implies\frac{m(n+p)+n(n+p)}{n(n+p)}\gt\frac{m}{n}\\\implies\frac{mn+mp+n^2+np}{n(n+p)}\gt\frac{m}{n}\implies\frac{n(m+p)+mp+n^2}{n(n+p)}\gt\frac{m}{n}$$
I can't seem to progress from here. I tried to find a way to force a cancellation with the $n$ in the denominator, but couldn't find a plausible approach. Any help?
Given your assumptions, write $$\frac {m+p}{n+p}-\frac mn=\frac{n(m+p)-m(n+p)}{(n+p)n}=\frac {(n-m)p}{(n+p)n}\gt 0$$
Consider $f(x) = \frac{m + x}{n + x}.$ This is an increasing function which can be verified by taking the derivative and seeing that it is everywhere positive. Hence,
$$ f(p) = \frac{m + p}{n + p} > f(0) = \frac{m}{n}.$$
Here is a short proof: as we have positive numbers, $$\frac{m+p}{n+p}>\frac mn\iff (\not m+p)n>(\not n+p)m\iff pn>pm\iff n>m.$$
This result can be extended to the reverse inequality. Here is a more expressive formulation:
Let $m,n,p\:$ be positive real numbers.
If $\frac mn <1$, we have $\;\frac mn <\frac{m+p}{n+p}<1$.
If $\frac mn >1$, we have $\;\frac mn >\frac{m+p}{n+p}>1$.
Alternative solution.
Let $p=am=bn$ for some real numbers $a,b>0$. Then $$\frac{m+p}{n+p}=\frac{m+am}{n+bn}=\frac{(1+a)m}{(1+b)n}=\frac{1+a}{1+b}\cdot\frac mn$$ so $$\frac{m+p}{n+p}>\frac mn\impliedby\frac{1+a}{1+b}>1\impliedby1+a>1+b\impliedby b<a$$ which is true since $m<n$ and $am=bn\implies\dfrac ba=\dfrac mn<1$. Therefore the original inequality is also true.
