Proving a function is positive. I'm looking to prove that the following inequality holds:
$\frac{(\ln x + 1)^{2} - \ln x}{(\ln x )^{2}} \geq \frac{x^{2}}{(1-x)^{2}} \quad \forall x \in (0,1) $ 
or equivalently that:
$f(x) =\frac{(\ln x + 1)^{2} - \ln x}{(\ln x )^{2}} -  \frac{x^{2}}{(1-x)^{2}} \geq 0\quad \forall x \in (0,1) $ 
Attempts so far have involved finding the limits of f(x) at 0 and 1 and trying to show that f(x) is decreasing over this interval. Also tried to show that f(x) has no turning points over the interval, but I can't seem to get anywhere. The latest idea was to try to find a function such that $ f(x) \geq g(x) \geq 0 $ over (0,1) but this just means I have to deal with 2 functions rather than 1.
Has anyone got any ideas how I might proceed? 
Thanks,
John.
 A: Let
$$f(x) =[(\ln x + 1)^{2} - \ln x](1-x)^{2}-(\ln x )^{2}x^{2},\quad \forall x \in (0,1). $$
Then the inequality is equivalent to $f(x)\ge0$ for $\forall x\in(0,1)$. Note
\begin{eqnarray}
f'(x)&=&3 x+\frac{1}{x}-2 \ln ^2x+2 \left(x+\frac{1}{x}-3\right) \ln x-4, \\
f''(x)&=&\frac{5 x^2-6 x+1+2 \left(x^2-2 x-1\right) \ln x}{x^2}. 
\end{eqnarray}
Let
$$ g(x)=\frac{5 x^2-6 x+1}{x^2-2 x-1}+2 \ln x, $$
then
$$ g'(x)=\frac{2 (x-1) \left(x^3-5 x^2-9 x-1\right)}{x \left(x^2-2 x-1\right)^2}. $$
For $x\in(0,1)$, it is easy to check that $x-1<0$ and $x^3-5 x^2-9 x-1<0$ and hence $g'(x)>0$. So $g(x)<g(1)=0$. Also note that $x^2-2x-1<0$ for $x\in(0,1)$ and
$$ \frac{x^2f''(x)}{x^2-2 x-1}=g(x)$$
and hence $f''(x)>0$ for $x\in(0,1)$. So $f'(x)<f'(1)=0$ for $x\in(0,1)$, which implies $f(x)>f(1)=0$ for $x\in(0,1)$.
A: Solve df/dx=0 and show that its positive at all those points if there are such points in the interval. Then check that its not negative at 0 or 1. But first think about and understand why this works.
