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 $l(p):=\ln\int_u^{u+1}x^p\,dx$.

Note that, 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.

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.

Here is a "more conceptual" and 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 $l(p):=\ln\int_u^{u+1}x^p\,dx$.

Note that, for each real $x>0$, $x^p$ is log convex in $p$ and hence $l$ is convex. Also, $l(0-)=0$. So, the "average slope" $\dfrac{l(p)}p$ of $l$ over the interval $(p,0)$ is increasing in $p<0$. 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.

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 $l(p):=\ln\int_u^{u+1}x^p\,dx$.

Note that, 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.