A stronger form of the weak $(1,1)$ inequality for the Hardy-Littlewood maximal function

Question

I am trying to show that for $f \in L^1(\mathbb R^d)$, if $f^*(x)$ is the Hardy Littlewood Maximal function, then the following inequality is satisfied:$$|\{x : f^*(x)> \alpha\}|\leq \dfrac{c}{\alpha}\int_{\{x:|f(x)|>\frac{\alpha}{2}\}}|f(x)|dx$$

I know that the following inequality holds:$$|\{x : f^*(x)> \alpha\}|\leq \dfrac{c}{\alpha}||f||_{L^1(\mathbb R^d)}$$ I've tried to separate the integral $\int_{\mathbb R^d}|f(x)|dx$ in the integral on the set $\{x:|f(x)|>\frac{\alpha}{2}\}$ and its complement but I couldn't arrive to anything.

I would appreciate hints or crucial steps to arrive to this inequality. Thanks in advance.

Answer 1

Following the hint by Christian Remling, let's write $f=f_1+f_2$ where $f_1=f\chi_{\{|f|\le \alpha/2\}}$. The maximal function is subadditive: $Mf\le Mf_1+Mf_2$. Also, $Mf_1\le \sup |f_1|\le \alpha/2$, which implies $$ \{Mf>\alpha \} \subset \{Mf_2>\alpha/2\} $$ Since $$|\{x : Mf(x)_2> \alpha/2\}|\leq \dfrac{c}{\alpha/2}\|f_2\|_{L^1(\mathbb R^d)}$$ the conclusion follows.