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Show, as elegantly as possible, that

$2p_0(1-p_0^r) > (p_0+\Delta)[1-(p_0+\Delta)^r] + (p_0 - \Delta)[1-(p_0-\Delta)^r]$

for $r \in \mathbb{R^+}$, $0 < p_0 - \Delta < p_0 + \Delta < (r+1)^{-\frac{1}{r}}$.

This is trivial for $r=1$, but I believe it should also hold for general $r > 0$. I would be interested in elementary proofs for that inequality -- if it is indeed true in general. Many thanks!

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up vote 0 down vote accepted

Your goal is to show that $u_r(p)\gt\frac12(u_r(p-\Delta)+u_r(p+\Delta))$ for every $0\lt\Delta\lt p$, where $u_r:p\mapsto p(1-p^r)$. Thus, if $u$ is concave, you are done. But $u_r''(p)=-(r+1)rp^{r-1}$ hence $u_r''(p)\lt0$ for every $p\gt0$, as soon as $r\gt0$. This shows that $u_r$ is indeed strictly concave.

Note: The upper bound $(r+1)^{-1/r}$ is irrelevant.

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Thanks for the excellent suggestions. Those proofs are certainly elementary enough, and the second is also quite elegant. – Martin Jul 24 '12 at 9:47
Elegant or not, there is only one proof in my post, hence I do not know what you are talking about. – Did Jul 24 '12 at 9:53
I meant to reply to both answers that were provided, not just yours - sorry for the confusion. – Martin Jul 24 '12 at 12:05
OK. Understood. – Did Jul 24 '12 at 12:57

This is almost true in general. If $r$ is even, the function on the right-hand side of the inequality is in fact constant, so the inequality does not hold.

This answer is not exactly elegant nor completely elementary; it just establishes the required result. I'll give a proof by calculus. First, note that for $\Delta = 0$, we have equality, so it is sufficient to show that $\Delta = 0$ is a local maximum in the given interval.

Define $f_{p_0}(\Delta) = (p_0 + \Delta)(1-(p_0+\Delta)^r) + (p_0 - \Delta)(1-(p_0 - \Delta)^r)$. Then $\Delta = 0$ is a maximum if $f^{\prime}_{p_0}(0) = 0$ and $f^{\prime\prime}_{p_0}(0) < 0$, where $f^{\prime}_{p_0}$ and $f^{\prime\prime}_{p_0}$ denote the first and second derivative of $f_{p_0}$.

We get $f^{\prime}_{p_0}(\Delta) = (r+1) ((p_0-\Delta)^r - (p_0+\Delta)^r)$, so $f^\prime_{p_0}(0) = 0$. Also, $f^{\prime\prime}_{p_0}(\Delta) = -r(r+1) ((p_0-\Delta)^{r-1} + (p_0+\Delta)^{r-1})$. Since $p_0 \pm \Delta > 0$, $p_0 > 0$ and hence, $f^{\prime\prime}_{p_0}(0) = -2r(r+1)p_0^{r-1} < 0$. Thus, we know that there is some interval around $\Delta = 0$ such that $f_{p_0}$ has a local maximum.

Analyzing $f'_{p_0}(\Delta) = 0$ gives $(p_0+\Delta)^r = (p_0-\Delta)^r$, or equivalently, $$\left( \frac{p_0+\Delta}{p_0-\Delta} \right)^r = 1$$ for the case that $\Delta \neq p_0$. Since there are at most two solutions to $x^r =1$ in the real numbers, namely $x = 1$ and $x = -1$, we need to consider two cases:

  1. $p_0 + \Delta = p_0 - \Delta$. Then $\Delta = 0$, which is the case we already know.
  2. $p_0 + \Delta = \Delta - p_0$. Then every $\Delta$ is a solution, but $p_0 = 0$.

In the first case, $f_{p_0}$ has exactly one extremal point, so the maximum is a global maximum on $\Delta < p_0$.

In the second case, $f_{p_0}$ is constant. Note that the second case is only possible if $r$ is even; in that case, the exception mentioned above applies.

The case $\Delta = p_0$ need not be considered, since $0 < p_0 - \Delta$ by assumption.

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Many thanks for the detailed answer! Martin – Martin Jul 24 '12 at 12:47

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