Convex functions in probability: $h(\alpha x_1 + (1 - \alpha)x_2) \le \alpha h(x_1) + (1 - \alpha)h(x_2)$

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I was studying probability and came across this statement:

If $h(x)$ is a convex function, that is, for $0 \le \alpha \le 1$

$$h(\alpha x_1 + (1 - \alpha)x_2) \le \alpha h(x_1) + (1 - \alpha)h(x_2)$$

for example $h(x) = x^2$ or $h(x) = e^x$, then for any $x_0$ there always exists a $b$ such that the line $h(x_0) + b(x - x_0)$ lies below $h(x)$, that is

$$h(x) \ge h(x_0) + b(x - x_0)$$

for any $x$. In particular, $h(X) \ge h(EX) + b(X - EX)$ for some $b$.

This entire thing looks like nonsense to me. The claims that it makes do not follow (are not implied by) the previous statements. For instance, it says that $h(\alpha x_1 + (1 - \alpha)x_2) \le \alpha h(x_1) + (1 - \alpha)h(x_2)$, but then it says that this implies that $h(x) \ge h(x_0) + b(x - x_0)$, which is at least unclear.

Could someone please clear this up, so that I can understand what point it's trying to make? Thank you.