%%visits: 2 if f is convex
f(average) ≤ Average(f)
$f(\sum _{k=1}^{n} x_k) \le \sum _{k=1}^{n} f(x_k)$
We can average this, every case of the function is less than, so if we sum each then divide by n, then we still have the same inequality.
f(g) + m(t-g) <= f(t)
“If I score better every test last year, then my average will always be better than last year”
jensens_inequalit := if f is convex, f(average) ≤ Average(f)