Proof: By Euclid

(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.
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