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.

$$\log x+\frac{1-x}{x} > 0
\iff \log x+\frac{1}{x} - 1> 0
\iff \frac{1}{x} -1 >  \log \frac 1x
$$
follows from
$$
\frac 1{1+t}< 1\implies \log(1+u) = \int_0^u \frac{dt}{1+t} < u
$$when $u>0.$
