Derivative of numerator greater than denominator implies quotient is increasing.

Question

Let $u, v$ be two non-negative functions satisfying $u'>v'>0$. Then I would like to show $$\left(\frac{u}{v}\right)'>0.$$ I'm not sure if this is actually even true, although it is intuitive and has worked for every example that I've tried. Here is my work so far:

\begin{align} \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}, \end{align} so it remains to show that $u'v-uv'>0$.

We can naively substitute the inequality $u'v-uv'>v'(v-u)$, but I believe $v-u>0$ is not required.

If someone can fill in what I'm missing then I would appreciate it.

Answer 1

You can try $u(x) = e^{2x}+1$ , $v(x)=e^x$ ; for negative $x$ ($x=-1$ for example), denominator is close to 0, when denominator increases by $0.001$, it is very big, and numerator is close to $1$, when numerator increases by $0.002$ or $0.003$, it is very small (proportionnaly), even if it is more than $0.001$.

Answer 2

A counterexample: $u(x)=2x+1,v(x)=x$. Then $u'=2>1=v'$ but clearly $u(x)/v(x)=2+1/x$ is decreasing in $x$ for $x>0$. Let's think about the intuition here:

---

Suppose you are making $\$200,000$/year and your friend is making $\$50,000$/year. Not bad, right? You're making quadruple what your friend is making! Now, let's say you got a raise of $\$100,000$/year and your friend got a raise of $\$50,000$/year. You got a bigger raise, right? But notice you are now making only triple what your friend is making. Why is that? Because your raise as a percentage of your income was smaller than that of your friend. That is,

$$3={\$200,000+\$100,000\over \$50,000 +\$50,000}={\$200,000\times 150\%\over \$50,000\times 200\%}<{\$200,000\over \$50,000}=4.$$

---

Thus, for a positive ratio to increase, you need the percent change in the numerator to exceed the percent change in the denominator. In continuous terms, the derivative of the log of a function captures the percent change in that function per unit change in the independent variable (in the limit as the change in the independent variable vanishes). Thus, for $u,v>0$, we can say $(\log u)'>(\log v)'$, or equivalently $u'/u>v'/v$, guarantees $u/v$ increases. The quotient rule also shows this since

$$\left({u \over v}\right)'={vu'-uv'\over v^2}=\left({u \over v}\right)\left( {u'\over u}-{v' \over v}\right),$$

or more compactly,

$$(\log (u/v))' =(\log u)'-(\log v').$$