A random variable X takes the maximum value of 80, and has a mean equal to 50. Give the best upper bound on P(X<=20).
So is it possible to use the Chebychev inequality here. Note that both values are 30 away from the mean. So is it correct to say that because P(X>=80) = 0 , then because P(X<=20) involves a similar calculation (because both values are 30 away from the mean), it would follow that P(X<=20) = 0 as well.
NOTE: We do not have the distribution of the RV.
You cannot use Chebyshev's inequality here, as you don't even know whether $X$ is square integrable (i.e., $X$ may not even have a standard deviation). (If you want, I can provide an example of such thing.)
So, among the three inequalities you list (Markov, Chebyshev, Chernoff), the only one applicable is the weakest, Markov. Using it on the non-negative r.v. $Y = 80-X$ which has expectation $30$, you get $$ \mathbb{P}\{X\leq 20\} = \mathbb{P}\{Y\geq 60\} \leq \frac{\mathbb{E}[Y]}{60} = \boxed{\frac{1}{2}}\,. $$
Note further that this bound cannot be improved without further assumptions, as shown by the following example: $X$ taking value $80$ with probability $\frac{1}{2}$, and value $20$ with probability $\frac{1}{2}$.