Power function inequality 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?
 A: 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.
A: [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.
A: 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.
