Jensen's Inequality (Real Analysis)/Corollary

Corollary to Jensen's Inequality: Real Analysis
Let $I$ be a real interval.

Let $\phi: I \to \R$ be a concave function.

Let $x_1, x_2, \ldots, x_n \in I$.

Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be real numbers, at least one of which is non-zero.

Then:
 * $\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$

For $\phi$ strictly concave, equality holds $x_1 = x_2 = \cdots = x_n$.

Proof
By Real Function is Concave iff its Negative is Convex, $-\phi: I \to \R$ is a convex function.

Therefore, we can apply Jensen's Inequality: Real Analysis with $-\phi$ to obtain:


 * $\ds -\map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le -\frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$

with equality for $-\phi$ strictly convex $x_1 = x_2 = \cdots = x_n$.

That is, for $\phi$ strictly concave, equality holds $x_1 = x_2 = \cdots = x_n$.

With that, we have established the equality case.

Multiplying through $-1$ in our inequality gives:


 * $\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$

as required.