For an arbitrary triangle, let $r$ = incircle radius and $R$ = circumcircle radius. Consider the ratio $r/R$? What is its maximum value? For an equilateral triangle this ratio is $1/2$. For very skewed triangles, this ratio approaches $0$. I am wondering if the maximum value is $1/2$.
This ratio is used as a measure of the quality of a triangulated surface mesh.
Your claim is correct.
We use the well-known area formulae $\Delta=rs=\frac{abc}{4R}=\sqrt{s(s-a)(s-b)(s-c)}$ to get $$\frac{r}{R}=\frac{4\Delta^2}{abcs}=\frac{4(s-a)(s-b)(s-c)}{abc}=\frac{4xyz}{(y+z)(z+x)(x+y)},$$ where $x=s-a, y=s-b,z=s-c$ are all positive. But $y+z\geq 2\sqrt{yz}$, $z+x\geq2\sqrt{zx}$, $x+y\geq2\sqrt{xy}$, so we get the result.
This result is called Euler's inequality. It is a corollary of the fact that $IO^2=R(R-2r)$, where $I$ is the incentre, $O$ the circumcentre.