Before the crucial step we make a few observations. 1) We observe that $L_{max}$ is strictly greater than $L_{min}$, as the lattice points create a triangle with rational area, however, the area of an equilateral triangle is irrational. 2) Both $L_{max}$ and $L_{min}$ are square roots of some positive integers. This follows due to the fact they can be expressed as $\sqrt{x^2+y^2}$, where $x,y$ are their respective lengths of projections on the coordinate axis. 3) While tampering with some small triangles, we can get a lower bound $k \ge 4$, which is achieved in the triangle $A(0;0), B(0;2), C(2;1)$ (it is easy to verify). Now we prove that $k=4$ satisfies the equation. Rewrite the inequality as $L^2_{max}-L_{max}L_{min} \ge \frac{1}{\sqrt{k}}$. Denote $L_{max}=\sqrt{a}$, $L_{min}=\sqrt{a-b}, \, b<a, \, a,b \in \mathbb{N}$. Then the inequality follows as this $$a-\sqrt{a(a-b)}\ge \frac{1}{2}$$But this is true, since $$a-\sqrt{a(a-b)}\ge a-\sqrt{a(a-1)}$$$$a-\sqrt{a(a-1)} \ge \frac{1}{2} \Longleftrightarrow \bigg(a-\frac{1}{2}\bigg)^2 \ge a(a-1)$$which is obviously true. $\, \square$