The implication $(1) \implies (2)$ is obvious, so suppose that $(2)$ holds. We proceed by induction on $N$. The base case of $N = 2$ is precisely the statement $(2)$, so as the induction hypothesis, suppose then that the claim holds for some integer $N \geq 2$. Let $x_1, \ldots, x_{N + 1} \in \mathbb{R}$ be real numbers, let $\lambda_1, \ldots, \lambda_{N + 1} \geq 0$ be unsigned real numbers with $\sum_{n = 1}^{N + 1} \lambda_n = 1$, and denote $\Lambda_N : = \sum_{n = 1}^N \lambda_n$. Then $\Lambda_N + \lambda_{N + 1} = 1$, $\sum_{n = 1}^N \frac{\lambda_n}{\Lambda_N} = 1$, and thus \begin{equation} \begin{split} f \left( \sum_{n = 1}^{N + 1} \lambda_n x_n \right) & = f \left[ \lambda_{N + 1} x_{N + 1} + \Lambda_N \left( \sum_{n = 1}^N \frac{\lambda_n}{\Lambda_N} x_n \right) \right] \\ & \leq \lambda_{N + 1} f(x_{N + 1}) + \Lambda_N f \left( \sum_{n = 1}^N \frac{\lambda_n}{\Lambda_N} x_n \right) \\ & \leq \lambda_{N + 1} f(x_{N + 1}) + \Lambda_N \sum_{n = 1}^N \frac{\lambda_n}{\Lambda_N} f(x_n) \\ & = \lambda_{N + 1} f(x_{N + 1}) + \sum_{n = 1}^N \lambda_n f(x_n) \\ & = \sum_{n = 1}^{N + 1} \lambda_n f(x_n) \end{split} \end{equation} The claim is now a consequence of R800: Proof by principle of weak mathematical induction. $\square$