Let $u,v,w\in\mathbb{R}^n$ be three points that are not collinear. We define $$ \triangle(u,v,w):=\{\alpha u+\beta v+\gamma w:\alpha+\beta+\gamma=1, \alpha,\beta,\gamma\geq 0\}, $$ $$ [u,v]=:= \{tu+(1-t)v:0\leq t\leq 1\}. $$ Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a function that is convex on $\triangle(u,v,w)$ and strictly convex on $\triangle(u,v,w)\setminus[u,v]$. My conjecture is that $f$ is strictly convex on the whole $\triangle(u,v,w)$. Could we prove or disprove this conjecture?
Remark. A function $f:C\rightarrow\mathbb{R}$, where $C\subset\mathbb{R}^n$ is a convex set, is called
convex if for every $u,v\in C, t\in (0,1)$ we have $$ f(tu+(1-t)v)\leq tf(u)+(1-t)f(v); $$
strictly convex if for every $u,v\in C, u\ne v, t\in (0,1)$ we have $$ f(tu+(1-t)v)< tf(u)+(1-t)f(v). $$
Note that strict convexity implies convexity and the vice versa is incorrect.
You can take $f(x,y) = x^2/y$, $u = (0,1)$, $v = (0,2)$ and $w = (1,2)$.
Note that the Hessian of $f$ is positive definite on $\triangle(u,v,w)\setminus[u,v]$ and positive semi-definite on $\triangle(u,v,w)$. But obviously, $f$ is not strictly convex on $[u,v]$.