Let $P$ is any point inside the triangle $ABC$ of side lengths $6,5,5$ units and $p_1,p_2,p_3$ be the lengths of perpendiculars drawn from $P$ to the sides of triangle. Find the maximum value of $p_1\cdot p_2\cdot p_3$
WLOG, I assumed the points $A(3,4)$ $B(0,0)$ $C(6,0)$
Now product of all three perpendiculars is $$\left|\frac{4a+3b-24}{5}\right|\cdot\left|\frac{4a-3b}{5}\right|\cdot|b|$$ where $(a,b)$ are coordinates of point $P$.
Now, how can I maximize the above expression$?$ The area of triangle is $12$ units.
I tried using $AM-GM$ inequality but in vain.
Should I go for Cauchy-Schwarz$?$
Any help is greatly appreciated.
Area of a triangle, A, in terms of side lengths $s_i$ and perpendicular lengths $p_i$, is $$A=\frac{\sum_{i=1}^3{p_i\cdot s_i}}{2}$$
AM-GM:$$\sqrt[3]{\prod_{i=1}^3{p_i\cdot s_i}}\le \frac{\sum_{i=1}^3{p_i\cdot s_i}}{3}$$
In this particular case the terms in AM-GM are not equal therefore the “$\le$ becomes “$\lt$”.
Therefore, $$3\sqrt[3]{\prod_{i=1}^3 p_i}\cdot \sqrt[3]{\prod_{i=1}^3 s_i}\lt\sum_{i=1}^3{p_is_i}=2A\Rightarrow \prod_{i=1}^3 p_i\lt\ \frac{(\frac{2}{3}A)^3}{\prod_{i=1}^3 s_i}$$
For A use Heron’s formula:$A=\sqrt{p\prod_{i=1}^3{(p-s_i)}}, p=\frac{s_1+s_2+s_3}{2}$
Numerically, $$\frac{(\frac{2}{3}A)^3}{\prod_{i=1}^3{s_i}}= \frac{\left(\frac{2}{3}\cdot 12\right)^3}{5\cdot5\cdot6}=\frac{256}{75}$$
$$\therefore p_1\cdot p_2 \cdot p_3 \lt \frac{256}{75}$$