Consider 3 points $A_1, A_2, A_3$ and a line $\varepsilon$, in the plane. If $A_iA'_i$, is the vertical segment to the line and $h_i, i=1, 2, 3$ are the heights of the triangle $A_1A_2A_3$ starting from the vertice $A_i$ then I would like to show: $$\sum A_iA'_i\ge \min\{h_i, i=1,2,3\}$$
I have tried to do that using elementary Geometry (I hope I did do any mistake) as follows:
First we see that if line $\varepsilon$ does not have common points with the triangle then we can always draw a line $\varepsilon'$ parallel to $\varepsilon$ which have at least one point in common with the triangle and the sum of distances is less than that of $\varepsilon$. Suppose now $\varepsilon$ crosses the triangle and the lines $\varepsilon$ and $A_2A_3$ meet at $O$. One of the feet $A'_i$ should be always inside the triange $A_1A_2A_3$, let's say $A'_1$ then the line $A_1A'_1$ meets line $A_2A_3$ to the point $A''_1$. So we have $\min\{h_i, i=1,2,3\}\leq h_1\leq A_1A''_1=A_1A'_1 + A'_1 A''_1\leq A_1A'_1 + A_3 A'_3\leq \sum A_iA'_i$,
as we wanted.
In case $\varepsilon//A_2A_3$ then again we have: $$\min\{h_i, i=1,2,3\} \leq h_1 \leq A_1''A_1 = A_1A'_1 + A'_1A''_1 = A_1A'_1 + A'_2A'_2 \leq \sum A_iA'_i$$
If the $A_i$ are colinear then heights are 0, so it is also true. My questions are:
- Is my "proof" correct or do you see any flaw?
- I have tried to prove the above with analytic geometry (see figure bellow) but I stuck.... providing a robust inequality, any advise would be appreciated
$$\sum A_iA'_i=\frac{|A\cdot 0+B\cdot 0+ C|+|A\cdot b+B\cdot 0+C|+|A\cdot c_1+B\cdot c_2+C|}{\sqrt{A^2+B^2}}\ge???$$