Parallelogram inside a parallelogram

Relevant wiki: Arithmetic Mean - Geometric Mean

Let $AD=BC = a$ and $AB=CD=b$ , then $K=2(a+b)$ . By cosine rule:

$\begin{aligned} \overline{EH}^2 & = \overline{AH}^2 + \overline{AE}^2 - 2 \cdot \overline{AH} \cdot \overline{AE} \cos \angle DAB \\ & = \frac {a^2}4 + \frac {b^2}4 - 2 \cdot \frac a2 \cdot \frac b2 \cos 60^\circ \\ \implies \overline{EH} & = \frac {\sqrt{a^2+b^2-ab}}2 \end{aligned}$

Similarly,

$\begin{aligned} \overline{HG}^2 & = \frac {a^2}4 + \frac {b^2}4 - 2 \cdot \frac a2 \cdot \frac b2 \cos 120^\circ \\ \implies \overline{HG} & = \frac {\sqrt{a^2+b^2+ab}}2 \end{aligned}$

Therefore, $k = \sqrt{a^2+b^2-ab} + \sqrt{a^2+b^2+ab}$ and

$\begin{aligned} \frac kK & = \frac {\sqrt{a^2+b^2-ab} + \sqrt{a^2+b^2+ab}}{2(a+b)} \\ & = \frac 12 \left(\sqrt{\frac {a^2+2ab+b^2-3ab}{a^2+2ab+b^2}} + \sqrt{\frac {a^2+2ab+b^2-ab}{a^2+2ab+b^2}} \right) \\ & = \frac 12 \left(\sqrt{1-\frac {3ab}{a^2+2ab+b^2}} + \sqrt{1-\frac {ab}{a^2+2ab+b^2}} \right) \\ & = \frac 12 \left(\sqrt{1-\frac 3{{\color{#3D99F6}\frac ab}+2+\color{#3D99F6}\frac ba}} + \sqrt{1-\frac 1{{\color{#3D99F6}\frac ab}+2+\color{#3D99F6}\frac ba}} \right) & \small \color{#3D99F6} \text{By AM-GM inequality } \frac ab + \frac ba \ge 2 \\ & {\color{#3D99F6} \le} \ \frac 12 \left(\sqrt{1-\frac 3{\color{#3D99F6}4}} + \sqrt{1-\frac 1{\color{#3D99F6}4}} \right) \\ & = \frac {1+\sqrt 3}4 \end{aligned}$

$\implies a+b+c = 1+3+4 = \boxed{8}$