Prove that $\ln{x} \leq \frac{4x}{x+4}$

Question

Can you prove that $\ln{x} \leq \frac{4x}{x + 4}$ for $\forall x > 0$? I can't. I tried using the inequality $\ln{x} \leq x - 1$ as follows:

$$\begin{align} x - 1 & \leq \frac{4x}{x+4} \\ x^2 + 3x - 4 & \leq 4x \\ x^2 - x - 4 & \leq 0 \\ x(x - 1) - 4 & \leq 0 \end{align}$$ so I can prove it for $0 < x < 3$, but I can't get any closer.

Thanks in advance.

Answer 1

The inequality is not valid for large values of $x$.

For example, let $x=100$, then $$\ln{x} \leq \frac{4x}{x + 4}$$

turns into $$4.605\le \frac {400}{104}=3.841$$

Which is not true.

Answer 2

Consider the function $$f(x)=\frac{4 x}{x+4}-\log (x)$$ When $x$ is large, by Taylor, $$f(x)=4-\frac{16}{x}-\log (x)+O\left(\frac{1}{x^2}\right)$$ and the zero of it is given in terms of Lambert function $$x=-\frac{16}{W\left(-\frac{16}{e^4}\right)}$$ Since the argument is small, we can use for the evaluation of $W(t)$ $$W(t)=t-t^2+\frac{3 t^3}{2}-\frac{8 t^4}{3}+\frac{125 t^5}{24}+O\left(t^6\right)$$ giving $x\approx 35.746$ while the exact solution would be $36.934$.

For $x=37$, $f(x)=\frac{148}{41}-\log (37)\approx -0.00116$ and then the inequality does not hold anymore.

Answer 3

Hint: In the interval $x>0$, the LHS is increasing without bound, while the RHS is increasing with the bound $4$ (the horizontal asymptote $y=4$).