For three points
![A,B,C](/media/m/6/0/1/6012c28979f41c54e9b40b9fc855aa34.png)
in the plane, we define
![m(ABC)](/media/m/9/1/7/917b62506e60f9effaefbc3f351741cb.png)
to be the smallest length of the three heights of the triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
, where in the case
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
,
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
,
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
are collinear, we set
![m(ABC) = 0](/media/m/a/b/2/ab2a1958d195a47f1843e4e61155d2f1.png)
. Let
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
,
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
,
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
be given points in the plane. Prove that for any point
![X](/media/m/9/2/8/92802f174fc4967315c2d8002c426164.png)
in the plane,
%V0
For three points $A,B,C$ in the plane, we define $m(ABC)$ to be the smallest length of the three heights of the triangle $ABC$, where in the case $A$, $B$, $C$ are collinear, we set $m(ABC) = 0$. Let $A$, $B$, $C$ be given points in the plane. Prove that for any point $X$ in the plane,
$$m(ABC) \leq m(ABX) + m(AXC) + m(XBC).$$