I have a random variable $X$ that arose from a Graph Theory problem, but I believe the details of the problem are not essential to this part, since the instructions explicitly say to use Markov's inequality. What might be important is that $0\leq X \leq m$ for some $m>1$. We have shown earlier that $E[X]\geq m/2$ and now are asked to use Markov's inequality to demonstrate that $P(X\geq 49m/100)\geq 1/51$.
I can't see how Markov's Inequality could be used to give a lower bound on a probability. I saw this question and answer: A lower bound on the probability that a variable is 3/2 times the expected value
I didn't follow some of the steps, but even so, in the end it appeared to me that the probability that was lower-bounded was just for the complement of the event in question. So as far as I can tell that solution would not help here.
Just for completeness I will write the inequality and do my best to try to use it:
$$P(X\geq 49m/100) \leq E[X]/(49m/100) \Rightarrow$$ $$49mp/100 \leq E[X]$$
where $p=P(X\geq 49m/100)$. I have heard rumor of a reverse Markov inequality in which I could compute
$$p \leq E[m-X]/(m-49m/100)$$ $$51mp/100 \leq E[m-X]$$
and I would imagine this further implies
$$51mp/100\leq m-E[X]$$ $$49mp/100\leq - E[X]$$ $$E[X]\leq 49mp/100$$
But clearly this result is paradoxical so I must not be fully understanding what the reverse Markov inequality really says, if it even is helpful here.