inequality proof involving natural logarithm

Question

Is there an elementary way for proving that $-\text{ln}(x)<\frac{1}{2x}$, for all $x \in ]0, \infty[$ ?

It's trivial for all $x \geq 1$, but what if $0<x<1$ ?

Answer 1

Let $$f(x)=\frac1{2x}+\ln x\implies f'(x)=-\frac1{2x^2}+\frac1x=\frac{2x-1}{2x^2}=0$$ for stationary points, giving $x=\dfrac12$. Now $$f''(x)=\frac1{x^3}-\frac1{x^2}\implies f''\left(\frac12\right)=4>0$$ so it is a global minimum. Since $f\left(\dfrac12\right)=0.306...>0$, we have that for all $x\in\mathbb{R}^+$, $f(x)>0$ so $$-\ln x<\frac1{2x}$$ as desired.

Answer 2

Note that

$$-\ln x<\frac{1}{2x}\iff\ln \frac1{x^2}<\frac1{x}$$

let $\frac1x=y$ for $y>0$

$$\iff \ln y^2<y\iff e^y>y^2$$

which is true since

• $f(y)=e^y-y^2 >0$ indeed $f(0)=1$ and $f'(y)=e^y-2y>0$ since
• $f''(y)=e^y-2=0$ for $y=\ln 2$ and $f'(\ln 2)=2-2\ln2>0$ and $f'''(y)=e^y>0$.