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Let the triangle be $ABC$, with area $[ABC]$. For any point $P$, let the distance to the sides be $p_a, p_b, p_c$ respectively. Connect the point $P$ to the 3 vertices. Considering the area of the triangles, we get:
$$ p_a a + p_b b + p_c c = 2[ABC]$$
[Note that you can use signed lengths to deal with the case where $P$ is outside of $ABC$]. Hence $p_a + p_b + p_c \geq \frac{ 2[ABC] } { \max(a,b,c) } $. It is easy to determine the equality cases, by considering the cases $a>b \geq c, a = b > c$ and $a = b =c$ separately.
Case 1: $a> b \geq c$. For equality to hold, $P$ must be the point $A$.
Case 2: $a=b > c$. For equality to hold, $P$ is any point on the line segment $AB$.
Case 3: $a=b=c$. For equality to hold, $P$ is any point within the triangle.