Prove that $ln(x)$ is concave by using the following definition for $ln(x)$

Question

Using the following definition: $$\ln(x) = \int_1^x \frac{1}{t} dt$$

Show that ln is concave. So basically what I need to show is that for $x,y \in \mathbb{R}^+, x \neq y$ and for some $t \in (0,1): \ln(tx+(1-t)y) < t\ln(x) + (1-t)\ln(y)$.

We have $$\ln(tx+(1-t)y) = \int_1^{tx+(1-t)y} \frac{1}{t} dt = \int_1^{tx} \frac{1}{t} dt + \int_{tx}^{(1-t)y} \frac{1}{t} dt$$

Assuming that F is the integral of $\frac{1}{t} dt$ we get: $F(tx)-F(1) + F((1-t)y)-F(tx) = F((1-t)y)-F(1)$ but this doesn't lead my anywhere. Any hints please?

Answer 1

If $F'$ is decreasing then $F$ is concave