2. $\mathbb EX > m_{X}$: $$\frac{1}{2}\leq \mathbb P \{X \leq m_{X} \} \leq \mathbb P \{-X+\mathbb EX \geq \epsilon \} \leq \mathbb P \{|X-\mathbb EX| \geq \epsilon \} \leq c_{1}e^{-c_{2}\epsilon^{2}}$$ So $$2c_{1} \geq e^{c_{2}\epsilon^{2}} \implies \ln 2c_{1} > 0$$

1. $\mathbb EX < m_{X}$: $$\frac{1}{2}\leq \mathbb P \{X \geq m_{X} \} \leq \mathbb P \{X-\mathbb EX \geq \epsilon \} \leq \mathbb P \{|X-\mathbb EX| \geq \epsilon \} \leq c_{1}e^{-c_{2}\epsilon^{2}}$$ So $$2c_{1} \geq e^{c_{2}\epsilon^{2}} \implies \ln 2c_{1} > 0$$

Actually the existence of such a $u$ should be proved and it is clear that, it does not conclude from $c_{1} \geq1$ and we should prove $\ln (2c_{1}) >0$. It can be proved like this: The problem in the case that $\mathbb EX=m_{X}$ is trivial. So we assume that $\mathbb EX \neq m_{X}$. Choose a real number $\epsilon$ such that $\epsilon < |\mathbb E-m_{X}|$. Now we consider two cases:

Thank you. The problem has a similar converse: Show that whenever for a median $m_{X}$ we know $$\mathbb P \{ |X-m_{X}| \geq t \} \leq c_{3}e^{-c_{4}t^{2}}$$ then $$\mathbb P\{|X-\mathbb EX|\geq t\}\leq 2c_{3}e^{-\frac{1}{4}c_{4}t^{2}}.$$