Proof $P(X=0)\le P(|X-\mu|\ge \mu)$ hold for a random variable $X\ge 0$?

Question

Is the following correct?

Let $X\ge 0$ be a non negative random variable and $0 \le E[X]=\mu$ the expectation value. Then $P(X=0)=P(X-\mu=-\mu)=P(-X+\mu=\mu)\le P(|-X+\mu|\ge \mu) = P(|-(X-\mu)|\ge \mu)=P(|X-\mu|\ge \mu).$

Do i need to consider anything about $P$ or $X$ or, if correct, is the above true as a general rule?

Answer 1

Note that $$ (X=0)\subset(|X-\mu|\geq \mu) $$ and the result follows from the fact that $A\subset B\implies P(A)\le P(B)$.