A question of minimising the expression.

Question

Let $P$ be any point inside $\Delta ABC$ whose distance from $\overline{BC}, \overline{AC}, \overline{AB}$ (or $a$, $b$ and $c$) be respectively $x$, $y$ and $z$. Then minimum value of $\frac{a}{x} + \frac{b}{y} + \frac{c}{z}$ is ?

Answer 1

By C-S $$(ax+by+cz)\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\geq(a+b+c)^2$$ Thus, $$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\geq\frac{(a+b+c)^2}{2S_{\Delta ABC}}.$$ The equality occurs for $$(ax+by+cz)||\left(\frac{a}{x},\frac{b}{y},\frac{c}{z}\right),$$ which says that $\frac{(a+b+c)^2}{2S_{\Delta ABC}}$ is a minimal value.

The inequality $$(ax+by+cz)\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\geq(a+b+c)^2$$ we can proof without C-S of course, but it's very ugly:

We need to prove that $$a^2+b^2+c^2+\frac{abx}{y}+\frac{aby}{x}+\frac{acx}{z}+\frac{acz}{x}+\frac{bcy}{z}+\frac{bcz}{y}\geq$$ $$\geq a^2+b^2+c^2+2ab+2ac+2bc$$ or $$ab\left(\frac{x}{y}+\frac{y}{x}-2\right)+ac\left(\frac{x}{z}+\frac{z}{x}-2\right)+bc\left(\frac{y}{z}+\frac{z}{y}-2\right)\geq0$$ or $$\frac{ab(x-y)^2}{xy}+\frac{ac(x-z)^2}{xz}+\frac{bc(y-z)^2}{yz}\geq0,$$ which is obvious.

Answer 2

prove that $$A=\frac{1}{2}(xa+yb+zc)$$ where $A$ is the area of the given triangle