Proof: By Euclid
(related to Proposition: 1.20: The Sum of the Lengths of Any Pair of Sides of a Triangle (Triangle Inequality))
 (So), (the sum of) $BA$ and $AC$ (is greater) than $BC$, (the sum of) $AB$ and $BC$ than $AC$, and (the sum of) $BC$ and $CA$ than $AB$.
 For let $BA$ have been drawn through to point $D$, and let $AD$ be made equal to $CA$ [Prop. 1.3], and let $DC$ have been joined.
 Therefore, since $DA$ is equal to $AC$, the angle $ADC$ is also equal to $ACD$ [Prop. 1.5].
 Thus, $BCD$ is greater than $ADC$.
 And since $DCB$ is a triangle having the angle $BCD$ greater than $BDC$, and the greater angle subtends the greater side [Prop. 1.19], $DB$ is thus greater than $BC$.
 But $DA$ is equal to $AC$.
 Thus, (the sum of) $BA$ and $AC$ is greater than $BC$.
 Similarly, we can show that (the sum of) $AB$ and $BC$ is also greater than $CA$, and (the sum of) $BC$ and $CA$ than $AB$.
 Thus, in any triangle, (the sum of) two sides taken together in any (possible way) is greater than the remaining (side).
 (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"