Let $D$ be an interior point of triangle $ABC$ such that $\angle ADB=\angle BDC=\angle CDA=\frac{2\pi}{3}$. Find the minimum $k$ such that $k(AB+BC+CA)\geq 2AD+\frac32BD+CD$ is always true.
The point $D$ is known as Fermat point and is also the point that minimizes the distance from the three vertices of the triangle. It can be assumed w.l.o.g that $AD\geq BD\geq CD$ to maximize the expression in the right hand side. For equilateral triangle $ABC$ with side length $1$, the left hand side is $3$ while the right hand side is $\frac{3\sqrt3}{2}$ which is about $2.6$, so $k\geq \frac{2}{\sqrt{3}}$.
It is best to put the Fermat point at the origin ($D\equiv O$) and assume $DA=x, DB=y, DC=z$.
By the cosine theorem $$ AB^2 = x^2+y^2+xy,\quad AC^2=x^2+z^2+xz,\quad BC^2=y^2+z^2+yz $$ so we just have to compute $$ k=\sup_{x,y,z > 0}\frac{2x+\frac{3}{2}y+z}{\sum_{cyc}\sqrt{x^2+y^2+xy}} $$ or, by exploiting homogeneity, $$ \inf_{\substack{x,y,z>0 \\ 2x+\frac{3}{2}y+z=1}}\sum_{cyc}\sqrt{x^2+y^2+xy} .\tag{1} $$ That can be tackled through standard techniques: we may find the stationary points on the boundary of the domain (a triangle in $\mathbb{R}^3$), then apply Lagrange multipliers to find stationary points inside the domain. As an alternative, we may eliminate a variable through $z=1-2x-\frac{3}{2}$ then study a two-variable function over a subset of $\mathbb{R}^2$.