Corollary to Markov's Inequality [closed]

Here is a particular case of Markov's Inequality that I failed to prove:

Let $X$ be a non-negative integer-valued random variable with $\mathbb{E}(X)\leq m$ then $$\mathbb{P}(X=0)\geq 1-m$$

Does anyone has an idea how to prove this? Thank you.

closed as off-topic by user21820, The Phenotype, zz20s, Ethan Bolker, NamasteFeb 22 '18 at 18:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, The Phenotype, zz20s, Ethan Bolker, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

$$m\geq\mathsf EX=\sum_{n=0}^{\infty}n\mathsf P(X=n)\geq\sum_{n=1}^{\infty}\mathsf P(X=n)=\mathsf P(X\geq1)=1-\mathsf P(X=0)$$
• To find it as corollary of Markovs inequality subtitute $a=1$ in $aP(X\geq a)\leq EX$ and work out. – drhab Feb 22 '18 at 12:05