Prove that the maximum volume of a triangular-base prism is $\sqrt{\frac{K^3}{54}}$ where $K$ is the area of three triangles containing a vertex $A$ Consider a prism with triangular base. The total area of the three faces containing a particular vertex $A$ is $K$. Show that the maximum possible volume of the prism is $\sqrt{\frac{K^3}{54}}$ and find the height of this largest prism.
I have no idea how to approach the problem. Please help. I know we need to use the properties of triangles and also the AM-GM inequality somewhere, but cannot put it together to solve the problem.
 A: WOLOG, consider the case $A$ is located at top of prism.
Let $A,B,C$ be the vertices of the triangle at top and $a,b,c$ be the lengths
of corresponding sides.
Let $\Delta$ be the area of triangle $\triangle_{ABC}$. Let $h$ and $V$ be the height and volume of the prism.
The given data is
$$K = \Delta + (b+c)h\quad\text{ and }\quad V = \Delta h$$
Notice $$\Delta \le \frac12 bc \le \frac18 (b+c)^2\quad\implies\quad (b+c) \ge \sqrt{8\Delta}$$
We have
$$K \ge \Delta + \sqrt{8\Delta} h
= \frac{V}{h} + \sqrt{8Vh} 
= \frac{V}{h} + \sqrt{2Vh} + \sqrt{2Vh}$$
Apply AM $\ge$ GM to rightmost expression, we get
$$K \ge 3\left(\frac{V}{h}\times \sqrt{2Vh} \times \sqrt{2Vh}\right)^{1/3}
=3(2V^2)^{1/3}$$
This is equivalent to the desired inequality 
$\displaystyle\;\sqrt{\frac{K^3}{54}} \ge V$.
Please note that this bound is tight. When $a : b : c : h = 2\sqrt{2} : 2 : 2 : 1$, triangle $\triangle_{ABC}$ is isosceles with a right angle at $A$.
Furthermore, the equality $\displaystyle\;\sqrt{\frac{K^3}{54}} = V$ is achieved.
