Let ''f'' be a function from an interval $I\; \backslash subseteq\; \backslash mathbb$ to $\backslash mathbb$. If ''f'' is convex, then for any three points ''x'', ''y'', ''z'' in ''I'',
:$\backslash frac\; +\; f\backslash left(\backslash frac\backslash right)\; \backslash ge\; \backslash frac\backslash left(f\backslash left(\backslash frac\backslash right)\; +\; f\backslash left(\backslash frac\backslash right)\; +\; f\backslash left(\backslash frac\backslash right)\; \backslash right).$
If a function ''f'' is continuous, then it is convex if and only if the above inequality holds for all ''x'', ''y'', ''z'' from $I$. When ''f'' is strictly convex, the inequality is strict except for ''x'' = ''y'' = ''z''.

Let ''f'' be a continuous function from an interval $I\; \backslash subseteq\; \backslash mathbb$ to $\backslash mathbb$. Then ''f'' is convex if and only if, for any integers ''n'' and ''k'' where ''n'' ≥ 3 and $2\; \backslash leq\; k\; \backslash leq\; n-1$, and any ''n'' points $x\_1,\; \backslash dots,\; x\_n$ from ''I'',
:$\backslash frac\; \backslash binom\; \backslash left(\; \backslash frac\; \backslash sum\_^f(x\_i)\; +\; nf\backslash left(\backslash frac1n\backslash sum\_^x\_i\backslash right)\; \backslash right)\backslash ge\; \backslash sum\_\; f\backslash left(\; \backslash frac1k\; \backslash sum\_^\; x\_\; \backslash right)$