Prove for that $(1 + \frac{x}{p})^p < (1 + \frac{x}{q})^q$ [without Bernoulli inequality or integrals]

Question

Prove for $x > 0$ and $0 < p < q$ that $$(1 + \frac{x}{p})^p < (1 + \frac{x}{q})^q$$

I think that binomial theorem might be of use in this exercise, but I'm not sure. I haven't been able to prove this on my own. Could you give me a hand with this?

EDIT: I'm not supposed to use integrals or Bernoulli inequality.

Answer 1

Since $\frac{q}{p}>1$, by using Bernoulli's inequality, we have \begin{align} 1+\frac{x}{p}=1+\frac{q}{p}\cdot\frac{x}{q}<\left(1+\frac{x}{q}\right)^{\frac{q}{p}}\quad\Longrightarrow\quad \left(1+\frac{x}{p}\right)^{p} < \left(1+\frac{x}{q}\right)^{q}. \end{align}

Answer 2

Hint This is equivalent to showing that the function $f(r) := \left(1 + \frac{x}{r}\right)^r$ is increasing. To compute the derivative, we need to take the logarithm, but since $\log$ is strictly increasing, it's anyway enough to show that $\log f$ is increasing, that is, that $\frac{d}{dr} \log f(r) > 0$. Computing gives $$\frac{d}{dr} \log f(r) = \log \left(1 + \frac{x}{r}\right) - \frac{x}{r \left(1 + \frac{x}{r}\right)} .$$ So, (rewriting in terms of the new variable $u := 1 + \frac{x}{r} > 1$) we just need to show that $$\log u > \frac{u - 1}{u}$$ for $u > 1$.

Additional hint One option is to rearrange this as $$u \log u - u + 1 > 0,$$ but we can recognize the l.h.s. as an antiderivative for $\log u$. After determining the appropriate constant we rewrite the desired inequality $$\int_1^u \log v \,dv > 0 ,$$ but the integrand is positive for $v > 1$, so the integral is positive for $u > 1$ as desired.