I was reviewing a bit of analysis, and for fun I thought to prove $f(x) = \ln(x)$ is concave on its domain just by using the definition of concave function. After having tried a bit, getting stuck in a passage, I tried to search for some hint and over here I found many similar questions but none was satisfying, because what I would like is:
No use of derivatives
No use of generalised/weighted AM-GM
No use of Jensen inequality
No use of the following Theorem:
If $f$ is continuous in $(a, b)$ such that$$f\left(\frac{x + y}{2}\right) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)$$for all $x, y \in (a, b)$, then $f$ is convex in $(a, b)$.
Here I am going to assume $\lambda \in (0, 1)$, for when $\lambda = 0$ or $1$, the starting inequality is trivially true.
Then starting with the definition, I wrote
$$\ln(\lambda x_1 + (1-\lambda)x_2) \geq \lambda \ln(x_1) + (1-\lambda)\ln(x_2)$$
It's all easy using, for example, the weighted AM-GM inequality. I thought, at most, of using the usual AM-GM inequality, but this lead me to a dead end:
$$\lambda x_1 + (1-\lambda)x_2 \geq 2\sqrt{\lambda x_1x_2}$$
hence
$$\ln(2\sqrt{\lambda x_1x_2}) = \frac{1}{2} \ln(4\lambda(1-\lambda)x_1x_2) = \frac{1}{2} \left( \ln(2\lambda x_1) + \ln(2(1-\lambda)x_2)\right)$$
Well from here it should go on by proving
$$\frac{1}{2} \left( \ln(2\lambda x_1) + \ln(2(1-\lambda)x_2)\right) \geq \lambda \ln(x_1) + (1-\lambda)\ln(x_2)$$
Or also
$$ \ln(2\lambda x_1) + \ln(2(1-\lambda)x_2) \geq 2\lambda \ln(x_1) + 2(1-\lambda)\ln(x_2)$$
But I'm not able to.
I could only think of unifying the left-hand side terms into a single log (that would be a step back), and writing the right terms as $(\ln(x_1^{2\lambda}) + \ln(x_2^{2-2\lambda})$
Thence again
$$\ln(4\lambda(1-\lambda)x_1x_2) \geq \ln(x_1^{2\lambda}x_2^{2-2\lambda})$$
And then by injectivity
$$4\lambda(1-\lambda)x_1x_2 \geq x_1^{2\lambda}x_2^{2-2\lambda}$$
And since $x_1,x_2 \neq 0$:
$$4\lambda(1-\lambda) \geq x_1^{2\lambda-1}x_2^{1-2\lambda}$$
Again, stuck.
Any hint?
