Inequality involving the distances from a fixed point in triangle to triangle sides and vertices

Question

Let $P$ be a point of an acute-angled triangle $ABC$, and $K,L,M$ be the projections of $P$ onto $BC,CA,AB$ respectively. Prove that $PK\cdot PL\cdot PM\le\frac{1}{8}\cdot PA\cdot PB\cdot PC$.

Answer 1

We'll prove that $$BC\cdot PA\geq AB\cdot PM+AC\cdot PL.$$ Indeed, let $CF$ is an altitude of $\Delta APC$ and $BG$ is an altitude of $\Delta APB.$

Thus, $$BC\cdot PA\geq(BG+CF)PA=2S_{\Delta APB}+2S_{\Delta APC}=AB\cdot PM+AC\cdot PL.$$ By the same we we can obtain: $$AC\cdot PB\geq AB\cdot PM+BC\cdot PK$$ and $$AB\cdot PC\geq BC\cdot PK+AC\cdot PL.$$ Thus, by AM-GM we obtain: $$AB\cdot AC\cdot BC\cdot PA\cdot PB\cdot PC\geq$$ $$\geq\left(BC\cdot PK+AC\cdot PL\right)\left( AB\cdot PM+BC\cdot PK\right)\left(AB\cdot PM+AC\cdot PL\right)\geq$$ $$\geq8\sqrt{AB^2AC^2 BC^2PK^2PL^2PM^2}=8AB\cdot AC\cdot BC\cdot PK\cdot PL\cdot PM$$ and we are done!