Prove $\frac{1}{\sqrt{x}}\geq \frac{\ln x}{x-1}$ I am trying to show that, for all $x>0:$
$$\frac{1}{\sqrt{x}}\geq \frac{\ln x}{x-1}$$
This inequality is closer than I expected. I have tried exponentiating, power series, and have achieved nothing. I would really appreciate some help. Below is a graph of the two functions for small $x:$

 A: Replacing $x$ with $x^2$ gives the more manageable inequality
$${1\over x} \ge {2\ln x\over x^2-1}\quad\text{ for all } x\gt0$$
to prove.  This can be settled by looking where the function 
$$f(x)=2\ln x + {1\over x}-x$$
crosses the $x$ axis, which it certainly does at $x=1$, but nowhere else, since
$$f'(x)={2\over x}-{1\over x^2}-1 = -\left(1-{1\over x}\right)^2$$
is always negative.
A: Here is one approach. Bring the right term to the left and make one fraction through a common denominator. The numerator has one zero for x=1.(Which is important!) The denominator is undefined for x=1. (The equality sign hereby is understood to be valid) For x between 0 and 1 as well as for x>1, determine the sign of the numerator and denominator. That should do it provided the given inequality is correct.
A: Would this approach work? 
For $x>1$
we need to show $\dfrac{\ln{x}}{x-1}\le\dfrac{1}{\sqrt{x}}$
$\sqrt{x}ln(x)\le x-1$
$ln(x)\le\sqrt{x}-\dfrac{1}{\sqrt{x}}$
which comes down to showing that $x\le e^{\sqrt{x}-\dfrac{1}{\sqrt{x}}}$
when x=1, we have equality
and when you differentiate the two, the following is true for x>1
$\dfrac{d}{dx}x \le \dfrac{d}{dx}e^{\sqrt{x}-\dfrac{1}{\sqrt{x}}} $
hence 
$x\le e^{\sqrt{x}-\dfrac{1}{\sqrt{x}}}$
so $\dfrac{\ln{x}}{x-1}\le\dfrac{1}{\sqrt{x}}$
 for x>1
A: Someone might have already mentioned this approach. But one thing you could do is bring everything to the left side and let the left side be $f(x$). So we have $f(x) \ge 0$. The idea here would be is to show that $f(x)$ begins positive, and then if you can show $f'(x) > 0$ for all $x > 0$, than $f(x)$ is a increasing function, and you would be done. 
$\frac{1}{\sqrt x} - \frac{ lnx}{x-1} \ge 0$ $\rightarrow$ $\frac{x-1-\sqrt xlnx}{\sqrt x (x-1)} \ge 0$ We can use calculus and show using L'hopitals rule as $x \rightarrow 0^-$ that $\sqrt xlnx \rightarrow 0$ so the numerator and denominator would be both negative as $x \rightarrow 0^-$ so if we let $f(x) = \frac{x-1-\sqrt xlnx}{\sqrt x (x-1)}$ we can say f(x) initially starts positive for any small positive x, and we just need to computationally confirm $f'(x) > 0$ for $x > 0$ and we would be done. 
