TITLE: Show that $f(x) = -\ln(x)$ is convex (WITHOUT using second derivative!)
QUESTION [5 upvotes]: In the lecture notes for a course I'm taking, the definition of a convex function is given as follows:
"a function $f$ is convex if, for any $x_1$ and $x_2$, and for any $\alpha$ $\in$ [0,1], $\alpha f(x_1) + (1-\alpha)f(x_2) \ge f(\alpha x_1 + (1-\alpha ) x_2)$" 
That is, if you draw a line segment between two points on the curve for this function, the function evaluated at any $x$ between $x_1$ and $x_2$ will be lower than the line segment.
Immediately after this definition is given, there is an exercise: "show that $f(x) = -\ln(x)$ is convex." Now, I happen to know that a function is convex if its second derivative is always greater than zero, so we can easily check the second derivative here to show that $-\ln(x)$ is convex (the second derivative is $\frac{1}{x^2}$, which is always greater than zero).
However, because of the way the exercise comes immediately after the "line segment" definition of convexity, without any mention of the second derivative test, I get the impression that it is possible to prove that $-\ln(x)$ is convex without using the second derivative test. I have attacked this problem many different ways, and I haven't been able to show convexity using only the line segment definition. Is this possible?

REPLY [2 votes]: Without the AGM nor the weighted AGM inequality. It suffices to consider the case $x> y$ and $a=\alpha \in (0,1).$ Take a fixed $y>0$ and a fixed $a\in (0,1)$ and for $x>0$ let $$g(x)=-a\log x- (1-a)\log y +\log (a x+(1-a)y).$$  We have $$g'(x)= \frac{dg(x)}{dx} = -\frac{a}{x} + \frac{a}{a x+(1-a) y} = \frac{a(1-a)(x-y)}{ax+(1-a)y}\cdot \frac {1}{x}.$$ Observe that $g'(y)=0$ and that $x>y\implies g'(x)>0.$ Therefore $$x>y\implies g(x)>g(y)=0.$$