Problem
The sum of the lengths of perpendiculars drawn from the three vertices of a triangle to their opposite sides is less than the sum of the lengths of the three sides of the triangle.
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Solution
As shown in the figure, regarding the right triangle $ABD$, $$AD<AB.$$ Regarding the right triangle $BCE$, $$BE<BC.$$ Regarding the right triangle $CAF$, $$CF<CA.$$ Therefore, $$AD+BE+CF<AB+BC+CA.$$
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Reference
Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.16