Suppose a coin gives heads $H$ with probability $\frac{1}{4}$. I flip it $100$ times and would like to estimate the upper bound on probability that I get more than or equal to $35$ heads.
If $X$ denotes number of heads in $100$ flips, then $E(X) = 25$ and $Var(X) = \frac{75}{4}$.
Applying Markov's inequality, I have that $P(X \geq 35) \leq \frac{25}{35} = \frac{5}{7}$. However applying Chebyshev's inequality, I have, since $\{X \geq 35\} \subset \{|X-25| \geq 10 \}$, the result:
$P(X \geq 35) \leq P(|X-25| \geq 10 ) \leq \frac{\frac{75}{4}}{100} = \frac{3}{16},$
which is smaller than $\frac{5}{7}$.
Isn't this contradictory? Markov gives a one sided bound, and this one sided bound is greater than the two-sided bound obtained by Chebyshev. Intuitively, one would expect that probability of getting 35 heads or more would be less than probability of getting either 35 heads or more, or 15 heads or less...