easy inequality to prove Prove that $\log_2(x)+\frac{1-x}{x} > 0$ 
I think the answer is easy but I've no clue how to go about it.

Take the exponents of both sides, it is equivalent to showing $xe^{\frac{1}{x}-1} > 1$. But $e^x = 1+x+\cdots > 1+x$ whenever $x$ is positive, and thus $xe^{\frac{1}{x}-1} > x(1+\frac{1}{x}-1) = 1$.
