Show that $f(x) = -\ln(x)$ is convex (WITHOUT using second derivative!)

It's expedient to apply the following result (see for example, Exercise $24$, Chapter $4$ of Rudin's Principles of Mathematical Analysis)

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 $(a, b)$.

Notice that $(a, b)$ can be extended to $(0, +\infty)$.

Here, if $x, y \in (0, + \infty)$, by AM-GM inequality, $$-\ln\left(\frac{1}{2}x + \frac{1}{2}y\right) \leq -\ln(\sqrt{xy}) = \frac{1}{2}(-\ln x) + \frac{1}{2}(-\ln y).$$ Since $-\ln x$ is continuous in $(0, +\infty)$, the proof is complete.

The Weighted AM-GM states

$$\lambda x+(1-\lambda)y\ge x^{\lambda}y^{1-\lambda}$$

Therefore, we have

$$\log(\lambda x+(1-\lambda)y)\ge \log(x^{\lambda}y^{1-\lambda})=\lambda \log(x)+(1-\lambda)\log(y)$$

Therefore the logarithm function is concave and its negative is convex.

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 $$x>y\implies g(x)>g(y)=0.$$