Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.
I would like to show, without calculus, that $AB + AC > BC$.
With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $B'C'$ below A, the slope of is zero. Call it D.
In the diagram,
$AB' = \int_y^z \sqrt{1+slope(AB')^2} dx$
$B'D = \int_z^w \sqrt{1+slope(B'C')^2} dx$
$AC' = \int_y^z \sqrt{1+slope(AC')^2} dx$
$C'D = \int_z^w \sqrt{1+slope(B'C')^2} dx$
The absolute value of the slope of $B'C'$ is decreasing from the slope of $AB'$ from B' to D, and negative but increasing (up to the absolute value of the slope of $AC'$) from D to C'
This is a general case of this question, which helped me when trying to place an upper limit on Pi.
