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Given is a triangle on points x,y,z in the plane. This triangle has two points a and b on different sides. I would like to show that the following inequality has to hold:

$\max \{d(b,x), d(b,y), d(b,z)\} + \max \{d(a,x), d(a,y), d(a,z)\} - d(b,a) \geq \min \{d(x,y), d(x,z), d(y,z)\}$

where d(u,v) denotes the euclidean distance between u and v. I actually expect the above statement to be true even if a and b are two arbitrary points outside of the triangle.

Does anybody have an idea how to approach this?

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What qualifies as "opposite sides" on a triangle? Each side is adjacent to the other 2. –  Mike Oct 26 '12 at 22:48
    
@Mike I replaced the word "opposite" with "different" –  andy Oct 26 '12 at 23:05
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Actually, it is true for all points inside the triangle (including the boundary). Assume that the line $ab$ intersects the side $xz$ when extended beyond $a$ and $yz$ when extended beyond $b$. Consider the quadrilateral $xaby$. We have $|xy|+|ab|\le|xb|+|ay|$ (the sum of opposite sides is not greater than the sum of diagonals). But the left hand side dominates $|ab|+\min(|xy|,|yz|,|xz|)$ while the right hand side is dominated by $\max(|ax|,|ay|,|az|)+\max(|bx|,|by|,|bz|)$. Outside the triangle the inequality may fail. Take any line $L$ and put $a$ and $b$ far away on that line on the opposite sides of the triangle. Then the difference on the left is almost the length of the projection of the triangle to $L$ but that dominates only the shortest altitude, not the shortest side.

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