Let $x$ and $p$ be real numbers with $x \ge 1$ and $p \ge 2$ . Show that $(x - 1)(x + 1)^{p - 1} \ge x^p - 1$ .
I recently discovered this result. I am sure it is known, but it is new to me. It is quite easy to prove if $p$ is an integer, even a negative one. I have a proof in the general case above, but it seems overly complicated. Can someone provide a simple demonstration?
We prove strict inequality for $x>1$ and $p>2$. Add $1$ to both sides and divide by $x^p$ to get an equivalent inequality that can be written as $$ \frac{x-1}{x} \left(\frac{x+1}{x}\right)^{p-1} + \frac1x \left( \frac1x \right)^{p-1} \geq 1. $$ Since $p > 2$ the function $f : X \mapsto X^{p-1}$ is strictly convex upwards. The left-hand side is a weighted average $$ \frac{x-1}{x} f\left(\frac{x+1}{x}\right) + \frac1x f\left( \frac1x \right) $$ of values of $f$, with positive weights and evaluated at different $X$'s. Hence by Jensen's inequality it strictly exceeds the value of $f$ at the corresponding weighted average of $X$'s, which is $$ f\left(\frac{x-1}{x} \cdot \frac{x+1}{x} + \frac1x \cdot \frac1x \right) = f(1) = 1, $$ QED.
The same argument shows that the inequality holds for $p<1$, and is reversed for $1 < p < 2$ because then $f$ is concave downwards.
Since the case $x=1$ is trivial, assume $x>1$ and divide both sides by $x-1$. It becomes $$(x+1)^{p-1}\ge \frac{x^p-1}{x-1} = 1 + x + x^2 + \cdots + x^{p-1}.$$ Expand the left side by the binomial theorem and compare it with the right side.
