# A geometric mean form of the Hermite-Hadamard inequality, for negative powers

The following inequality appeared in the analysis of a random approximation algorithm: $$\int_u^{u+1} x^p\ \mathrm{dx} \leq \sqrt{u^p(u+1)^p}\text{, for } -1\leq p\leq 0, u\geq 1.$$

This resembles the well-known Hermite-Hadamard inequality for convex functions $$\int_a^b f(x)\ \mathrm{dx} \leq (b-a)\frac{f(a)+f(b)}{2}$$ but with the right-hand side being the geometric mean, rather than the arithmetic mean $$(u^p+(u+1)^p)/2$$.

Computationally, this inequality appears to hold, but I have been unable to find a proof. Is this a known result? If so, is there a broader class of convex functions for which it holds?

The inequality in question is a particular case (with $$v=u+1$$) of the inequality $$\int_u^v x^p\, dx \le(v-u)u^{p/2}v^{p/2}\tag{1}$$ for $$v\ge u>0$$, where without loss of generality (wlog) $$p\in(-1,0]$$. By the homogeneity in $$(u,v)$$, wlog $$u=1$$, and then (1) can be rewritten as $$g(v):=g_p(v):=v^{p+1}-1-(p+1)(v-1) v^{p/2}\le0\tag{2}$$ for $$v\ge1$$.

Note that
$$g'(v)=\tfrac12\, (p+1) v^{p/2-1}h_v(p),\tag{3}$$ where $$h_v(p):=2 v (v^{p/2}-1)+p-p v,$$ so that $$h_v(p)$$ is convex in $$p$$. Also, $$h_v(0)=0$$ and $$h_v(-1)=-(\sqrt v-1)^2\le0$$. So, $$h_v\le0$$ and hence $$g'\le0$$, which implies that $$g$$ is decreasing, from $$g(1)=0$$. Thus, (2) follows.

One may also note that $$h_v(-2)=0$$, so that $$h_v(p)\le0$$ for all $$p\in[-2,0]$$, whence, by (3), $$g'=g'_p\ge0$$ for $$p\in[-2,-1)$$, which implies that $$g$$ is increasing, from $$g(1)=0$$. So, for $$p\in[-2,-1)$$, inequality (2) switches the direction, and (1) continues to hold -- because (2) was obtained by multiplying both sides of (1) by $$p+1$$. (The extension to $$p\in[-2,-1)$$ was previously suggested by Alapan Das.)

This inequality fails to hold e.g. when $$p=-3$$ and $$u=1$$. (Then the left-hand side of the inequality is $$3/8$$, whereas the right-hand side is $$2^{-3/2}$$, so that the ratio of the left-hand side to the right-hand side is $$\sqrt{9/8}>1$$.)

With the additional condition that $$p\ge-1$$, the inequality holds, in fact for all real $$u>0$$. Indeed, without loss of generality $$-1. Let $$R(u)$$ denote the ratio of the left-hand side to the right-hand side. Then $$R_1(u):=R'(u)S(u)=\frac{u^{-p} (u+1)^p \left((2-p) u-p+2 u^2\right)}{(p+2) u+2 u^2}-1,$$ where $$S(u):=\frac{2 (p+1) u^{1-p/2} (u+1)^{1+p/2}}{(p+2) u+2 u^2}>0,$$ so that $$R'(u)$$ has the same sign as $$R_1(u)$$. Moreover, $$R'_1(u)=\frac{p (p+1) (p+2) u^{-p-2} (u+1)^p}{(p+2 u+2)^2}\le0.$$ Also, $$R_1(\infty-)=0$$. So, $$R_1\ge0$$ and hence $$R(u)$$ is increasing in $$u>0$$, to $$R(\infty-)=1$$.

So, $$R(u)\le1$$ for $$u>0$$, as desired.

• Thank you. I'm really only interested in the case when $p\geq -1$, but my code was misbehaving and had indicated the inequality held for all nonpositive $p$. Statement corrected. Commented Mar 30, 2021 at 18:56
• Inequality for $R_1^\prime$ inverted Commented Mar 30, 2021 at 20:12
• @username : No, it is not. Remember that $-1<p\le0$. Commented Mar 30, 2021 at 20:15
• I stand corrected Commented Mar 30, 2021 at 20:16

Substituting $$x\rightarrow \frac{x}{u}=z$$ and taking $$1/u=y\leq 1$$ we define a function $$f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$$ where $$r=-p, t=t/2$$.

Now, $$f(0)=0$$ and $$f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$$

Again, as $$(1+y)>1$$ we define $$f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$$

Hence, $$f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$$. So, we carry on doing this and get $$f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)>0$$

Here, $$T_{n+1}=(1+t)T_{n}-t^{n}$$.

$$T_3=(1+t-t^2)>1$$ as $$t<0.5 \Rightarrow T_4=(1+t)T_3-t^4=T_3+T_3t-t^4>1$$ and hence, $$T_n >1$$ $$\forall n \in \mathbb N$$

For, $$n>>0, t^n \rightarrow 0$$ as $$0\leq t \leq 1/2$$ $$f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$$

[For, large $$n>>0$$ as $$f_n(0)=\epsilon_n>0$$ we expect to find some $$N_0$$ such that $$t^ny(1+y)^t<\epsilon_n$$ for all $$n>N_0$$ and $$0\leq y \leq 1$$]

Hence, proved. So, it must hold for $$-2\leq p \leq 0$$

Here is a "more conceptual" and more general proof:

For real $$u>0$$ and $$p<0$$, let
$$m(p):=\Big(\int_u^{u+1}x^p\,dx\Big)^{1/p},$$ so that $$\ln m(p)=\frac{l(p)}p,$$ where $$l(p):=\ln\int_u^{u+1}x^p\,dx$$.

For each real $$x>0$$, $$x^p$$ is log convex in $$p$$ and hence $$l$$ is convex -- see e.g. this. Also, $$l(0-)=0$$. So, the "average slope" $$\dfrac{l(p)}p$$ of $$l$$ over the interval $$(p,0)$$ is increasing in $$p<0$$; here one can use e.g. the special-case l'Hospital-type rule for monotonicity, Proposition 4.1. So, $$m$$ is increasing on $$(-\infty,0)$$. So, $$m(p)\le m(-2)\quad\text{for}\quad p\in(-\infty,-2]$$ and $$m(p)\ge m(-2)\quad\text{for}\quad p\in[-2,0).$$

The latter inequality is just another way to write the inequality in question, which thus holds for all $$p\in[-2,0]$$ and all real $$u>0$$.

For $$p\in(-\infty,-2]$$, the direction of the inequality in question changes to the opposite.