I want to show that $|f(x)|\le(Mf)(x)|$ at every Lebesgue point of $f$ if $f\in L^1(R^k)$

I want to show this, where $Mf$ is a maximal function, and I have attain $$Mf(x)-|f(x)|=\sup_{0\le r\le\infty}\frac{1}{B(x,r)}\int_{B(x,r)}(|f(y)|-|f(x)|)dm(y)$$ and I have no idea how to show that RHS is $\ge 0$. My guess is to use $$||x|-|y||\le|x-y|.$$

By the definition of Lebesgue point, triangle inequality, and the definition of $Mf$, $$\begin{split} |f(x)| &= \left|\lim_{r\to 0}\frac{1}{B(x,r)}\int_{B(x,r)}f(y)\,dm(y)\right| \\ &\le \lim_{r\to 0}\frac{1}{B(x,r)}\int_{B(x,r)}|f(y)|\,dm(y) \\ &\le \sup_{r> 0}\frac{1}{B(x,r)}\int_{B(x,r)}|f(y)|\,dm(y) \\ &=Mf(x) \end{split}$$