# 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?

• I like the phrasing of your question; it suggests curiosity, politeness, and the desire to learn. Unfortunately, there is lack of motivation, and I can't tell if we are being asked to solve a homework problem. I recommend asking this on math.stackexchange, with motivation provided. Also, I suspect this question is outside the scope of this forum. Gerhard "Ask Me About System Design" Paseman, 2011.10.19 – Gerhard Paseman Oct 19 '11 at 17:38
• @Gerhard: According to the OP's profile, and, as corroborated by the math genealogy database, it seems highly unlikely this is a homework question. :) (That said, I agree it's still off-topic.) – cardinal Oct 19 '11 at 19:00
• Also, this question has already appeared on math.SE: math.stackexchange.com/questions/71758/… – cardinal Oct 19 '11 at 19:03
• @Richard: I have posted an answer to the math.SE version. I'm not sure if you were expecting something simpler than that. Cheers. – cardinal Oct 19 '11 at 19:21
• If the poster had added what method he used to show it, or what he was going to do with the result, it would be a more appropriate question for MathOverflow. If my advisor or Terry Tao or gowers or anyone else had posted the same question with the same wording, I would still ask for motivation and remark that it looked like a homework question to me. That said, I agree it is a nice problem and hope a good answer is forthcoming on some forum. Gerhard "Ask Me About System Design" Paseman, 2011.10.19 – Gerhard Paseman Oct 19 '11 at 19:28

## 2 Answers

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.

• Very nice! I wonder how robust is this approach in general. – Seva Oct 19 '11 at 20:29
• @Seva: Thanks! What do you mean by "robust" in this context? – Noam D. Elkies Oct 20 '11 at 0:35
• I mean something of the following sort: given a "random inequality", what are the chances that it can be proved by applying Jensen's inequality to an appropriately selected function. Pretty high, I guess... – Seva Oct 20 '11 at 8:18
• Beautiful argument--thanks. As cardinal observed over at math.stackexchange, you don't need Jensen, just convexity. – Richard Hevener Oct 20 '11 at 19:12
• @Seva: what probability measure are you using to select a "random inequality"? ;-) Jensen / convexity is a versatile tool for proving inequalities of mathematical interest; the application here feels somewhat unusual but I'm sure it's not the strangest use I've seen of the technique. – Noam D. Elkies Oct 21 '11 at 5:37

Since the case $x=1$ is trivial, assume $x>1$ and divide both sides by $x-1$. It becomes $$(x+1)^{p-1}\ge \frac{x^p-1}{x-1} = 1 + x + x^2 + \cdots + x^{p-1}.$$ Expand the left side by the binomial theorem and compare it with the right side.

• As Richard observed, the case when p is integral is easily done. Can you provide something almost as nice for nonintegral p? Gerhard "Ask Me About System Design" Paseman, 2011.10.20 – Gerhard Paseman Oct 21 '11 at 4:32
• Yeah, I realised that after posting. Oops. – Brendan McKay Oct 21 '11 at 7:04
• If you started this way with rational p (and perhaps letting x=y^(pq) for a suitable integer pq), you might redeem the approach, introduce some cool combinatorics, extend the result to real p by continuity, and provide an extension to the result in a fashion similar to what Noam Elkies did. I would do it myself, but after rewriting I get (y^r +1)^(pq) >= something to the power r which needs more cool combinatorics to handle than I can muster currently. Gerhard "The Buck Passes Here, Sometimes" Paseman, 2011.10.23 – Gerhard Paseman Oct 23 '11 at 20:30