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.

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.)

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
$$g'(v)=\tfrac12\, (p+1) v^{p/2-1}h_v(p),$$ 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.

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.