Proving a particular case of AM-GM graphically

Question

Sketch a graph of $y = \ln x$ and hence explain why $$\frac{\ln a + 2 \ln b}{3} \le \ln\frac{a + 2b}{3}$$

I understand that $\ln $ is a strictly increasing function and $x > y \implies \ln x > \ln y$. With this, it is trivial to show that since

$$\frac{a + b + b}{3} \ge \sqrt[3]{a\cdot b \cdot b}$$

then it follows that

$$\ln\frac{a + 2b}{3} \ge \ln\sqrt[3]{ab^2} = \frac{\ln a + 2 \ln b}{3}$$

However, this does not make use of the graph at all. In fact, it is highly likely that this solution is inappropriate: the next part of the question is to derive the AM-GM inequality for this particular case (Hence derive an ineqauality connecting $ab^2$ and $a + 2b$).

Answer 1

we have $f(x)=\ln(x)$ by differentiating we get $$y''=-\frac{1}{x^2}<0$$ thus we can apply Jensen's inequality $$\frac{f(a)+f(b)+f(b)}{3}\geq f\left(\frac{a+b+b}{3}\right)$$

Answer 2

This is related with the convex of the function $\ln x$.