We can extend BA past A into a straight line.
There exists a point D such that DA = CA.
Therefore, \angle ADC = \angle ACD because isosceles triangle have two equal angles.
Thus, \angle BCD > \angle BDC by Euclid's fifth common notion.
Since \triangle DCB is a triangle having \angle BCD greater than \angle BDC, this means that BD > BC.
But BD = BA + AD, and AD = AC.
Thus, BA + AC > BC.
A similar argument shows that AC + BC > BA and BA + BC > AC.