The proof proceeds by mathematical induction on $n$. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$ $\map P 1$ is true, as this just says: :$\displaystyle \map \phi {\frac {\lambda_1 x_1} {\lambda_1} } \le \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}$ {{begin-eqn}} {{eqn | l = \map \phi {\frac {\lambda_1 x_1} {\lambda_1} }       | r = \map \phi {x_1}       | c =  }} {{eqn | r = \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}       | c =  }} {{end-eqn}} trivially. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^r \lambda_k x_k} {\sum_{k \mathop = 1}^r \lambda_k} } \le \frac {\sum_{k \mathop = 1}^r \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^r \lambda_k}$ where $\lambda_1, \lambda_2, \ldots, \lambda_r \ge 0$ are real numbers, at least one of which is non-zero. Then we need to show: :$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k x_k} {\sum_{k \mathop = 1}^{r + 1} \lambda_k} } \le \frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^{r + 1} \lambda_k}$ where $\lambda_1, \lambda_2, \ldots, \lambda_{r + 1} \ge 0$ are real numbers, at least one of which is non-zero. === Induction Step === This is our induction step: {{WLOG}}, assume that $\lambda_r$ and $\lambda_{r + 1}$ are non-zero. Otherwise, the statement reduces to the induction hypothesis. Define: :$y := \dfrac {\lambda_r x_r + \lambda_{r + 1} x_{r + 1} } {\lambda_r + \lambda_{r + 1} }$ Then: {{begin-eqn}} {{eqn | l = \map \phi {\frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k x_k} {\sum_{k \mathop = 1}^{r + 1} \lambda_k} }       | r = \map \phi {\frac {\sum_{k \mathop = 1}^{r - 1} \lambda_k x_k + \paren {\lambda_r + \lambda_{r + 1} } y} {\sum_{k \mathop = 1}^{r - 1} \lambda_k + \paren {\lambda_r + \lambda_{r + 1} } } } }} {{eqn | o = \le       | r = \frac {\sum_{k \mathop = 1}^{r - 1} \lambda_k \map \phi {x_k} + \paren {\lambda_r + \lambda_{r + 1} } \map \phi y} {\sum_{k \mathop = 1}^{r - 1} \lambda_k + \paren {\lambda_r + \lambda_{r + 1} } }       | c = by the induction hypothesis }} {{eqn | o = \le       | r = \frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^{r + 1} \lambda_k}       | c = {{Defof|Convex Real Function}} }} {{end-eqn}} Note that in the case that $\phi$ is strictly convex, equality holds {{iff}} $x_1 = x_2 = \cdots = x_r = x_{r + 1}$. So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{>0}: \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$ {{qed}} {{Namedfor|Johan Ludwig William Valdemar Jensen|cat = Jensen}} Category:Real Analysis Category:Inequalities Category:Jensen's Inequality (Real Analysis) l6ou3qzrwkkf4ri0es47nvpdnz3z14p