Need help on a short proof Prove that on $(0,1)$ that $$-\ln(x)/(1-x) < 1/\sqrt{x}$$.
I tried looking at the derivative but didn't help.
 A: Hint:
For $x \in (0,1)$, you have $1 - x \gt 0$, so
$$-\frac{\ln(x)}{1-x} \lt \frac{1}{\sqrt{x}} \iff -\ln(x) \lt \frac{1 - x}{\sqrt{x}} = \frac{1}{\sqrt{x}} - \sqrt{x} \tag{1}\label{eq1A}$$
I believe using this to create a function to check its values and derivative should be considerably easier than using the original inequality.
Update: Since the OP has stated below they already previously determined this inequality, here is how to go further. You can create a function
$$f(x) = \ln(x) + \frac{1}{\sqrt{x}} - \sqrt{x} \tag{2}\label{eq2A}$$
Note $f(1) = 0$. Also, you have
$$\begin{equation}\begin{aligned}
f'(x) & = \frac{1}{x} - \frac{1}{2x^{3/2}} - \frac{1}{2x^{1/2}} \\
& = \frac{1}{2x^{3/2}}(2x^{1/2} - 1 - x) \\
& = -\frac{1}{2x^{3/2}}(1 - \sqrt{x})^2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, the derivative is negative for $x \in (0,1)$, so the value of $f(x)$ is strictly decreasing to $f(1) = 0$. This shows that $f(x) \gt 0$ for $x \in (0,1)$, which confirms that \eqref{eq1A} holds.
A: Note that, for $x\in (0,1)$,
$$\frac1{\sqrt{x}}+\frac{ \ln x}{1-x}=\frac1{1-x}\left( \frac{1}{\sqrt x}-\sqrt x + \ln x\right)=\frac1{1-x}\int_x^1 \left( \frac1{2t^{3/2}} + \frac1{2t^{1/2}} - \frac1t\right)dt$$
$$=\frac1{1-x}\int_x^1  \frac{(1-t^{1/2})^2}{2t^{3/2}}>0$$ 
Thus, 
$$-\frac{ \ln x}{1-x}< \frac1{\sqrt{x}}$$
