For $k$ to make sense, we should assume that $\|u\|_p\ne0$. Let $l(x)$ and $r(x)$ denote the left- and right-hand sides of your displayed inequality. Then, by Tonelli's theorem, $$\int dx\,l(x)=\int dx\int dz\, 1\{|z-x|<1\}|\nabla u(z)|^p \\ =\int dz\,|\nabla u(z)|^p\int dz\, 1\{|z-x|<1\} =v_n\|\nabla u\|_p^p\le v_n,$$ where $v_n$ is the volume of the unit ball in $\mathbb R^n$. Similarly, $$\int dx\,r(x) =kv_n\|u\|_p^p>v_n.$$ So, $$\int dx\,l(x)<\int dx\,r(x)$$ and hence $$l(x)<r(x)$$ for at least one $x$, as desired.

For $k$ to make sense, we should assume that $\|u\|_p\ne0$. Let $l(x)$ and $r(x)$ denote the left- and right-hand sides of your displayed inequality. Then, by Tonelli's theorem, for any real $p>0$ $$\int dx\,l(x)=\int dx\int dz\, 1\{|z-x|<1\}|\nabla u(z)|^p \\ =\int dz\,|\nabla u(z)|^p\int dz\, 1\{|z-x|<1\} =v_n\|\nabla u\|_p^p\le v_n,$$ where $v_n$ is the volume of the unit ball in $\mathbb R^n$. Similarly, $$\int dx\,r(x) =kv_n\|u\|_p^p>v_n.$$ So, $$\int dx\,l(x)<\int dx\,r(x)$$ and hence $$l(x)<r(x)$$ for at least one $x$, as desired.