Proof: We use Cauchy induction to prove this lemma. By assumption, when $n=2$, the inequality holds, so the inequality holds for $n$ being a power of 2. Therefore, we only need to prove that if the inequality holds for $n=k+1$ (where $k \in \mathbb{N}, k \geq 2$), then it also holds for $n=k$. In fact, suppose the inequality holds for $n=k+1$. Let $x = x_1 + x_2 + \cdots + x_k$ and $x_{k+1} = \frac{x}{k}$. By the induction hypothesis, we have
$$f(x_1) + f(x_2) + \cdots + f(x_k) + f\left(\frac{x}{k}\right) \geq (k+1) f\left(\frac{x + \frac{x}{k}}{k+1}\right) = (k+1) f\left(\frac{x}{k}\right)$$

Proof complete.

The result above can be directly obtained from Jensen's inequality, because by definition, every convex function $f$ satisfies $\left(t = \frac{1}{2}\right)$
$$f(x) + f(y) \geq 2 f\left(\frac{x + y}{2}\right)$$

Obviously, if we change the condition $f(x) + f(y) \geq 2 f\left(\frac{x + y}{2}\right) \forall x, y \in [a, b]$ to $f(x) + f(y) \leq 2 f\left(\frac{x + y}{2}\right) \quad \forall x, y \in [a, b]$, then the inequality will change direction
$$f(x_1) + f(x_2) + \cdots + f(x_n) \leq n f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right)$$