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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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    Since $x \geq 1$, you can divide out $x-1$ from both sides of the inequality...2011-10-11
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    Sure, but, what do you get when you divide, say, $x^{\pi}-1$ by $x-1$?2011-10-11
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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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    The right side is correct only for integer values of $p$.2011-10-11
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    The above proof is valid for positive integers. As I indicated in the statement of the problem, there is no difficulty in the integer case.2011-10-11
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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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    Nice answer. Perhaps I am missing something a little subtle, but it seems you don't need to appeal to Jensen's inequality here. The definition of strict convexity would seem to suffice.2011-10-20
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    Beautiful argument--thanks. I agree with cardinal that convexity is all that is required, not Jensen.2011-10-20
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    You're welcome, and thanks for the check-mark. Re convexity vs. Jensen -- they're equivalent, and I learned the inequality some decades ago as "Jensen's inequality"; perhaps by now it's been assimilated thoroughly enough to be regarded as the *definition* of convexity?2011-10-20
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    Thanks for your follow-up comment. Out of curiosity, what were you taking as the definition of *strict convexity*?2011-10-21
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    Since $f$ is twice differentiable it's certainly enough that $f''$ is strictly positive. In general it's enough for $f''$ not to vanish on an interval of positive length. For a continuous function that need not be differentiable, it can be convenient to use the definition $f((x+y)/2) < (f(x) + f(y))/2$ for $x \neq y$ (e.g. to prove that a positive power series is logarithmically convex).2011-10-21
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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