Power function inequality

Question

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?

Answer 1

[Copied from my answer to the same question on mathoverflow, where "Cardinal" noted the question's previous appearance here]

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.

Answer 2

When $x=1$, it is trivial. Otherwise, it is equivalent to showing $(x + 1)^{p - 1} \ge x^{p-1}+x^{p-2}+\cdots+1$. The binomial expansion of the left hand side .....

Editted In case $p$ is not integer, you'd better prove $f(x)=(x - 1)(x + 1)^{p - 1}- x^p +1$ is nonnegative, this is not difficult.

Answer 3

Let $f(x) = (x-1)(x+1)^{p-1} - x^p + 1$ and note that $f(1) = 0$.

Now, $$ f'(x) = (p-1)(x-1)(x+1)^{p-2} + (x+1)^{p-1} - p x^{p-1} \>, $$ and rewriting the last two terms, we get $$ f'(x) = (p-1)(x-1)(x+1)^{p-2} + x^{p-1}\left(\left(1+\frac{1}{x}\right)^{p-1} - p\right) \> . $$

By Bernoulli's inequality, $$ \left(1+\frac{1}{x}\right)^{p-1} - p \geq 1+(p-1)/x - p = -x^{-1}(p-1)(x-1). $$

Hence, $$ f'(x) \geq (p-1)(x-1)((x+1)^{p-2} - x^{p-2}) \geq 0 \>, $$ and so $f(x)$ is nondecreasing for $x \geq 1$, which yields the desired result.