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?



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




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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.)


