Let $(X,\mu)$ be a measure space. Then for each $p$ between $0$ and $1$,we have $$\lVert f_{1}+\cdots+f_{m}\rVert_{L^{p,\infty}}^{p}\le \frac{2-p}{1-p}\sum_{j=1}^{m}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}$$
Then we denote the distribution function as $\omega_{|f|}(\alpha)=\mu\bigg(\{x\in X:|f(x)|>\alpha\}\bigg)$
Here is my details to this proof (I edited):
First of all, if there were any one $\lVert f_{j}\rVert_{L^{p,\infty}}\,\,$ is infinity,then our conclusion holds easily.
So we may assume each $\lVert f_{j}\rVert_{L^{p,\infty}}<\infty$
For each $\alpha>0$,we observe that the following,
\begin{align} &\alpha\mu\bigg(\{x\,:|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,,\max_{1\le j\le m}|f_{j}(x)|\le\alpha \}\bigg)\\ &=\int_{\{x\,:|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,,\,\max_{j}|f_{j}(x)|\le\alpha \}}\alpha&\,d\mu(x)\\ &\le\int_{\{x\,:|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,,\,\max_{j}|f_{j}(x)|\le\alpha \}}|f_{1}(x)+\dots+f_{m}(x)|&\,d\mu(x)\\ &\le\int_{\{x\,:|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,,\,\max_{j}|f_{j}(x)|\le\alpha \}}|f_{1}(x)|+\dots+|f_{m}(x)|&\,d\mu(x)\\ &\color{red}\le\int_{\,{\{x\in X\, :\,\max_{j}|f_{j}(x)|\le\alpha\}}} |f_{1}(x)|+\cdots+|f_{m}(x)|\,\,d\mu(x)\\ &\color{red}=\sum_{j=1}^{m}\int_{\{x\in X\, : \,\max_{j}|f_{j}(x)|\le\alpha\}} |f_{j}(x)|\,\,d\mu(x)\\ &\le \sum_{j=1}^{m}\int_{\{x\, : \,|f_{j}(x)|\le\alpha\}}|f_{j}(x)|\,d\mu(x)\\ &\leq\sum_{j=1}^{m}\frac{\alpha^{1-p}}{1-p}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}\\ \end{align}
Then we yields a consequence that for each $\alpha>0$,
$$\alpha^p\mu\bigg(\{x\in X :|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,,\max_{1\le j\le m}|f_{j}(x)|\le\alpha \}\bigg) \le\sum_{j=1}^{m}\frac{1}{1-p}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}$$
Therefore we combine the previous result that for each $\alpha>0$,one has \begin{align} &\alpha^p\mu\bigg(\{x\in X :|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,\}\bigg)\\ &\le\alpha^p\mu\bigg(\{x\in X :|f_{1}(x)+\cdots+f_{m}(x)|>\alpha\,,\max_{1\le j\le m}|f_{j}(x)|\le\alpha \}\bigg)+\alpha^{p}\omega_{\max_{j}|f_{j}|}(\alpha)\\ &\le \sum_{j=1}^{m}\frac{1}{1-p}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}+\alpha^{p}\omega_{\max_{j}|f_{j}|}(\alpha)\\ &\le \sum_{j=1}^{m}\frac{1}{1-p}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}\,\,+\bigg\lVert \max_{1\le j\le m}|f_{j}|\bigg\rVert_{L^{p,\infty}}^{p}\\ &\le\sum_{j=1}^{m}\frac{1}{1-p}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}\,\,+\sum_{j=1}^{m}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}\\ &=\bigg(1+\frac{1}{1-p}\bigg)\sum_{j=1}^{m}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}\\ &=\frac{2-p}{1-p}\sum_{j=1}^{m}\lVert f_{j}\rVert_{L^{p,\infty}}^{p} \end{align}
To the end of our proof when we take sup for every $\alpha>0$ at left hand side.
Well,if you have the time,please check this for validity,or just ignore it that it's okay . Any suggestion and valuable advice would be greatest appreciated.A lot of thanks for patient reading.
The line that \begin{align*} \int_{|f_{1}+\cdots+f_{m}|>\alpha,\max|f_{j}|\leq\alpha}|f_{1}|+\cdots+|f_{m}|d\mu\leq\sum_{i=1}^{m}\int_{|f_{i}|>\alpha/m,\max|f_{j}|\leq\alpha}|f_{i}|d\mu \end{align*} does not hold. Rather, the correct one should be \begin{align*} \int_{|f_{1}+\cdots+f_{m}|>\alpha,\max|f_{j}|\leq\alpha}|f_{1}|+\cdots+|f_{m}|d\mu\leq\sum_{i=1}^{m}\int_{|f_{i}|>\alpha/m,\max|f_{j}|\leq\alpha}|f_{1}|+\cdots+|f_{m}|d\mu \end{align*}
However, you are not far away to the goal. Here's the crucial step. \begin{align*} \alpha\mu(|f_{1}+\cdots+f_{m}|>\alpha,\max|f_{j}|\leq\alpha)&=\alpha\mu_{E}(|f_{1}+\cdots+f_{m}|>\alpha)\\ &\leq\int|f_{1}+\cdots+f_{m}|d\mu_{E}\\ &=\int_{E}|f_{1}+\cdots+f_{m}|d\mu\\ &=\int_{\max|f_{j}|\leq\alpha}|f_{1}+\cdots+f_{m}|d\mu\\ &\leq\sum_{i=1}^{m}\int_{\max|f_{j}|\leq\alpha}|f_{i}|d\mu\\ &\leq\sum_{i=1}^{m}\int_{|f_{i}|\leq\alpha}|f_{i}|d\mu\\ &\leq\sum_{i=1}^{m}\dfrac{\alpha^{1-p}}{1-p}\|f_{i}\|_{L^{p,\infty}}^{p}\\ &=\alpha^{1-p}\cdot\dfrac{1}{1-p}\sum_{i=1}^{m}\|f_{i}\|_{L^{p,\infty}}^{p}, \end{align*} where $E=\{x\in X: \max|f_{j}(x)|\leq\alpha\}$, $\mu_{E}(A)=\mu(A)\chi_{E}$, rearranging we get \begin{align*} \alpha^{p}\mu(|f_{1}+\cdots+f_{m}|>\alpha,\max|f_{j}|\leq\alpha)\leq\dfrac{1}{1-p}\sum_{i=1}^{m}\|f_{i}\|_{L^{p,\infty}}^{p}. \end{align*}