Analytical proof of the inequality $p^n (1-p\ln p)<1$ where $0<p<1$

Question

Simulations with Mathematica suggest to me that $$p^n (1-p\ln p) <1\quad\text{for } p \in (0,1) \text{ and }n \ge 0$$

Have you got any hints on how I can derive an analytical proof of this result?

Many thanks in advance!

Answer 1

Hagen von Eitzen rightly pointed out in his comment that your inequality becomes false for $n\ge 0$. However, if you restated your problem for $n>0$, then your conjecture from Mathematica simulations holds.

To show it analytically, you could start by defining the function $$f(p)=p^n-p^{n+1}\ln p$$ for $p\in\left]0;1\right[$ and $n>0$. (Notice that you could extend this interval to also include 1, as it is not a forbidden value, contratry to 0).

From there, simply compute the first derivative with respect to $p$: $$f'(p)=np^{n-1}-(n+1)p^n\ln p-p^n$$ Now, you want to study the sign of $f'(p)$ on the interval. Let's break it into two parts:

1. On $\left]0;1\right[$, notice that $p^{n-1}>p^n$, such that $np^{n-1}-p^n>0$;
2. Also, since $0<p<1$, then $p^n>0$ and $\ln p<0$, such that $-(n+1)p^n\ln p>0$.

Combining both lines:

$$np^{n-1}-(n+1)p^n\ln p-p^n>0$$ $$\therefore f'(p)>0$$ Since $f'(p)$ is strictly positive on the interval, $f(p)$ must be strictly increasing on the same interval. The final step is to compute the limit of the function at each boundary: $$\lim_{p\to 0}f(p)=0$$ $$\lim_{p\to 1}f(p)=f(1)=1$$ In conclusion, $f$ is strictly increasing on $\left]0;1\right[$, and bounded above by $f(1)=1$, such that your (modified) conjecture is true.