I want to show that $-\log(x^2) \geq 2(1-x)$ for $x \in (0,1]$, where $\log$ is the logarithm to base 2. How can I do that?
I tried to make an estimate by first bringing the minus to the other side and then exponientiating, getting that the above is equivalent to $x^2 \leq e^{\ln(2)2(1-x)}$ for $x \in (0,1]$. Then I wanted to cut the series expansion of $e^{\ln(2)2(1-x)}$ and show that what I get is larger than $x^2$, but I don't know how to argue that the residual that I cut away is positive.
Hint: $\ln (x^2) = 2 \ln x$. let $f(x) = \ln x - x + 1$. Find the global maximum of this function, and develop an inequality based on that.