how to prove this inequality? Given $x>0$, $y>0$ and $x + y =1$, how to prove that $\frac{1}{x}\cdot\log_2\left(\frac{1}{y}\right)+\frac{1}{y}\cdot\log_2\left(\frac{1}{x}\right)\ge 4$ ?

Hint 1: Rewrite this inequality as:
$$-x\log_2 x - (1-x)\log_2 (1-x) \geq 4 x (1-x)$$
Both sides of the inequality define concave functions on the interval $[0,1]$. Plot them. Can you show that the graph of the second is always lying below the graph of the other? 
