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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?

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    Somehow I skipped the "$p$ is real" bit... :D2011-10-20

3 Answers 3

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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.

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    I'm still waiting for the proof in case _p_ is a real number > 2.2011-10-13
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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.

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    Fine argument--thanks.2011-10-20
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[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.

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    Thanks. I was actually curious if you might have been taking continuity plus midpoint convexity as your starting point. Not to belabor the point any further, but, that said, if you appeal to the second-derivative equation for strict convexity, then you get the full version f(t x + (1-t)y) < t f(x) + (1-t) f(y) for free. It essentially falls out of (one of) the standard proof(s) using the MVT. If you use midpoint convexity as your point of departure, then you need the extra continuity assumption (as you state) and then, perhaps, something like Jensen becomes necessary. :)2011-10-21