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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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  • $\begingroup$ Since $x \geq 1$, you can divide out $x-1$ from both sides of the inequality... $\endgroup$ Commented Oct 11, 2011 at 17:14
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    $\begingroup$ Sure, but, what do you get when you divide, say, $x^{\pi}-1$ by $x-1$? $\endgroup$ Commented Oct 11, 2011 at 22:12
  • $\begingroup$ Somehow I skipped the "$p$ is real" bit... :D $\endgroup$ Commented Oct 20, 2011 at 0:46

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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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  • $\begingroup$ 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. $\endgroup$
    – cardinal
    Commented Oct 20, 2011 at 14:04
  • $\begingroup$ Beautiful argument--thanks. I agree with cardinal that convexity is all that is required, not Jensen. $\endgroup$ Commented Oct 20, 2011 at 19:01
  • $\begingroup$ 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? $\endgroup$ Commented Oct 20, 2011 at 20:44
  • $\begingroup$ Thanks for your follow-up comment. Out of curiosity, what were you taking as the definition of strict convexity? $\endgroup$
    – cardinal
    Commented Oct 21, 2011 at 2:16
  • $\begingroup$ 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). $\endgroup$ Commented Oct 21, 2011 at 5:32
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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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  • $\begingroup$ Fine argument--thanks. $\endgroup$ Commented Oct 20, 2011 at 19:02
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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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  • $\begingroup$ The right side is correct only for integer values of $p$. $\endgroup$
    – robjohn
    Commented Oct 11, 2011 at 18:55
  • $\begingroup$ 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. $\endgroup$ Commented Oct 11, 2011 at 22:21
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    $\begingroup$ I'm still waiting for the proof in case p is a real number > 2. $\endgroup$ Commented Oct 13, 2011 at 15:10

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