If $f$ is a convex function on the interval $[a,b]$, for each $\left\{x_{k}\right\}_{{k=1}}^{n}\in[a,b]$ and each $\left\{\mu_{k}\right\}_{{k=1}}^{n}$ with $\mu_{{k}}\geq 0$ one has:

A common situation occurs when $\mu_{1}+\mu_{2}+\cdots+\mu_{n}=1$; in this case, the inequality simplifies to:

where $0\leq\mu_{k}\leq 1$.

If $f$ is a concave function, the inequality is reversed.

Example:
$f(x)=x^{2}$ is a convex function on $[0,10]$. Then

A very special case of this inequality is when $\mu_{k}=\frac{1}{n}$ because then

that is, the value of the function at the mean of the $x_{k}$ is less or equal than the mean of the values of the function at each $x_{k}$.

There is another formulation of Jensen’s inequality used in probability:
Let $X$ be some random variable, and let $f(x)$ be a convex function (defined at least on a segment containing the range of $X$). Then the expected value of $f(X)$ is at least the value of $f$ at the mean of $X$:

With this approach, the weights of the first form can be seen as probabilities.