Consider a general triangle $ABC$ and the three inequalities given by the theorem.

Here, it will be shown that $AC+AB>BC.$ The other two inequalities can be proved following the same procedure. Start by extending $\overline{AB}$ to the left of $A.$ Then, consider a point $D$ on this line such that $AD=AC.$

In $△ADC,$ the sides $\overline{AD}$ and $\overline{AC}$ are congruent. This means that by the Isosceles Triangle Theorem, the angles opposite them are congruent angles. Therefore, $\angle D\cong \angle DCA,$ which means that $m\angle D=\angle DCA.$

Notice that $\angle DCB$ is made of $\angle DCA$ and $\angle ACB.$ Therefore, the measures of these three angles can be related thanks to the Angle Addition Postulate. From this equation, $m\angle DCB$ is greater than $m\angle DCA.$ Now consider $△DCB.$ From the previous inequality and the Triangle Larger Angle Theorem, the side opposite $\angle DCB$ is longer than the side opposite $\angle D.$ The length of $\overline{DB}$ is equal to the sum of the lengths of $\overline{DA}$ and $\overline{AB}.$ This is because of the Segment Addition Postulate. Substituting the corresponding sum into the left-hand side results in the following inequality. Recall that $D$ was plotted so that $AD=AC.$ Therefore, substitute $AC$ for $DA$ into the last inequality. Notice that the last inequality is the desired one. Consequently, the proof is complete.