TITLE: Prove that $f$ is convex in an interval given an inequality with determinant
QUESTION [12 upvotes]: Today,I found a interesting problem:

if 
  $$\begin{vmatrix}
\cos{x}&\sin{x}&f(x)\\
\cos{y}&\sin{y}&f(y)\\
\cos{z}&\sin{z}&f(z)
\end{vmatrix}\ge 0$$
  for all $x,y,z$ of an open interval $I$ for which $x<y<z<x+\pi$.
show that:
  $f(x)$ is convex on $I$

REPLY [6 votes]: Rewrite the equation as
$$
\begin{align}
0
&\le f(x)\sin(z-y)-f(y)\sin(z-x)+f(z)\sin(y-x)\\
&=(f(z)-f(y))\sin(y-x)+f(y)\sin(y-x)(1-\cos(z-y))\\
&-(f(y)-f(x))\sin(z-y)+f(y)\sin(z-y)(1-\cos(y-x))\tag{1}
\end{align}
$$
Divide by $(z-y)(y-x)$ and take the double limit $\liminf\limits_{x\to y^-}\liminf\limits_{z\to y^+}$
$$
\begin{align}
0
&\le\lim_{x\to y^-}\liminf_{z\to y^+}\frac{f(z)-f(y)}{z-y}\frac{\sin(y-x)}{y-x}\\
&+\lim_{x\to y^-}\lim_{z\to y^+}f(y)\frac{\sin(y-x)}{y-x}\frac{1-\cos(z-y)}{z-y}\\
&+\lim_{x\to y^-}\lim_{z\to y^+}f(y)\frac{\sin(z-y)}{z-y}\frac{1-\cos(y-x)}{y-x}\\
&-\limsup\limits_{x\to y^-}\lim_{z\to y^+}\frac{f(y)-f(x)}{y-x}\frac{\sin(z-y)}{z-y}\tag{2}
\end{align}
$$
which becomes
$$
\limsup\limits_{x\to y^-}\frac{f(y)-f(x)}{y-x}\le\liminf_{z\to y^+}\frac{f(z)-f(y)}{z-y}\tag{3}
$$
Inequality $(3)$ implies that $f$ is convex. The rest follows from convexity.