Prove that $f$ is convex in an interval given an inequality with determinant

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 $I$

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.