If $(x,y,z)$ be the lengths of perpendiculars drawn from any interior point P of a triangle ABC on the sides BC, CA and AB respectively, then find the minimum value of $x^2+y^2+z^2$. The sides-lengths of the triangle $ABC$ being $a$, $b$ and $c$.
My try
I have found that the area of the triangle would be $\frac{ax+by+cz}{2}$. So that can form a constraint equation. Can I go ahead and minimise $x^2+y^2+z^2$ with the constraint ${ax+by+cz}=K$(some constant) using Lagrange's multiplier method? Would that be the correct way?
By C-S $$(x^2+y^2+z^2)(a^2+b^2+c^2)\geq(ax+by+cz)^2=4S^2.$$ Thus, $$x^2+y^2+z^2\geq\frac{2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4}{4(a^2+b^2+c^2)}.$$ The equality occurs for $(x,y,z)||(a,b,c)$, which says that we got a minimal value.
We'll prove that there exist $P$ inside the triangle for which $x:y:z=a:b:c$.
Let $L\in AB$ such that $\frac{S_{\Delta LBC}}{S_{\Delta LAC}}=\frac{a^2}{b^2}$ and $M\in AC$ such that $\frac{S_{\Delta MBC}}{S_{\Delta MAC}}=\frac{a^2}{c^2}$.
Now, let $BM\cap CL=\{K\}$ and easy to see that for this $K$ we get an equality case.