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

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