show that $0.5 - \frac{ln(x)}x > 0$

Question

I try to show that $$ \frac{1}{2} - \frac{ln(x)} x > 0 $$ on $x \in (0, \infty)$ ,

any help would be appreciated, but the simplest way will be the best

Answer 1

We can rearrange $$0.5−\frac{\ln(x)}{x}>0$$ to be $$0.5>\frac{\ln(x)}{x}$$ multiply the x over (which is always going to be positive) $$\frac{1}{2}x>\ln(x)$$ make them each exponents of for $e^x$ $$e^{\frac{1}{2}x}>e^{\ln(x)}$$ which is $$e^{\frac{1}{2}x}>x$$ $$\sqrt{e}\cdot e^x>x$$ and exponential functions always grow faster than polynomial functions, and in this case, is always greater than x.

If you didn't like that solution, you could use calculus: Take the derivative of $\frac{\ln(x)}{x}$

$$\frac{x\cdot\frac{1}{x}-\ln(x)\cdot1}{x^2}$$

and find where it equals goes from positive to negative: $1-\ln(x)=0$ @ $x = e$

so we know that because the slope goes positive to negative, we have a maximum @ $x = e$.

$$\ln(e)/e=\frac{1}{e}$$ which is our maximum of this function, which is less than 0.5, so we know that $$0.5>\frac{\ln(x)}{x}$$ and thus $$0.5−\frac{\ln(x)}{x}>0$$

Answer 2

Since $x>0$, the inequality is equivalent to $$ x-2\ln x>0 $$ Find the minimum of the function $$ f(x)=x-2\ln x $$ using the fact that $$ f'(x)=1-\frac{2}{x} $$