# ${P(|X|\geq \lambda)\leq f(\lambda)}$

I have to prove a bound of the form $$P(|X|\geq \lambda)\leq f(\lambda)\quad (1),$$where $f$ denotes some upper bound function and $X$ is a complex random variable.

My question is: I know a bound on the real and imaginary part of the random variable, i.e. $$P(|\Re X|\geq \lambda)\leq f_1(\lambda)\\P(|\Im X|\geq \lambda)\leq f_2(\lambda),$$how can I use these bounds to prove the bound $(1)$, i.e. to construct $f$ ?

We have: $$\max\{|\Re X|,|\Im X|\}\leq|X|=\sqrt{|\Re X|^2+|\Im X|^2}\leq |\Re X|+|\Im X|.$$ Then $$|X|\geq\lambda\implies |\Re X|\geq\lambda/2\quad\text{or}\quad|\Im X|\geq\lambda/2$$ so that $$\Pr(|X|\geq\lambda)\leq\Pr[(|\Re X|\geq\lambda/2)\cup(|\Im X|\geq\lambda/2)]\leq f_1(\lambda/2)+f_2(\lambda/2).$$ Thus, $f(\lambda)\equiv f_1(\lambda/2)+f_2(\lambda/2)$ works.