Let $u \in H^1(\Omega) \cap C(\Omega)$ ($\Omega \subset R^n$ a smooth and bounded domain) a nonnegative function. Let $B(x,R) \subset \overline{ B(x,R) } \subset \Omega $ a ball. Suppose that
$$\frac{2}{r \sigma_{n-1}} \int_{\partial B(x, r/2)} u \geq 5, $$
where ${\sigma_{n-1}} : = \int_{\partial B(0, 1)} 1$.
The text that I am reading says that "by Chebyshev inequality we can find $y \in \partial B(x,r/2)$ such that $u(y) \geq \frac{5r}{2}$".
I don't know which version of the inequality the author is using... I dont know how to apply the versions that I know. Someone could help me to understand his argument?
Thanks in advance
What you have is is that the average value of $u$ on $\partial B$ (which is $\frac{\int_{\partial B(x,r/2) } u}{\sigma_{n-1}}$) is greater than or equal to $\frac{5r}{2}$.
Since the average value of the function $u$ is at least $\frac{5r}{2}$, there must exist a point on the domain of integration such that the value is at least $\frac{5r}{2}$.