Does the existence of one angle in a triangle imply the triangle inequality?

Question

It seems intuitive that, given three lengths of $a,b$ and $c$ and given that an angle opposite the side $c$ of $C$ exists such that $-1 \leq \cos(C)\leq 1 \implies -1\leq \dfrac{a^2+b^2-c^2}{2ab} \leq 1$, we should be able to prove that these lengths obey the triangle inequality. I have also graphed this function $(\cos(C))$ to find that whenever it is within these bounds it obeys the triangle inequality at least for a finite range of inputs. Of course this is not a proof, so I have also attempted to prove this, but after going through the algebra, I find $(a-b)^2 \leq c^2 \leq (a+b)^2$. Now, I have always struggled with understanding how to manipulate inequalities, so I am unsure how to proceed here. What exactly do these two inequalities imply about the square roots of each side? When you take the square root of both sides of an inequality, you obviously can not treat it like you would with equality, as in that there are two solutions of $LHS \leq \pm{RHS}$ as this is self contradictory. How would I get all the solutions required for the triangle inequality then? As in: $a\leq b+c, b\leq a+c$ and $c\leq a+b$. What are the logical steps in solving for the square roots of inequalities?

Answer 1

From $(a-b)^2 \leq c^2 \leq (a+b)^2$, we can deduce that $|a-b|\leq |c|\leq |a+b|$ by taking positive square roots (taking negative square roots would reverse order). Since $c,a+b\geq 0$, it follows that $c\leq a+b$ and $|a-b|\leq c$. The latter gives $a-b\leq c \implies a\leq b+c$ and $b-a\leq c \implies b\leq a+c$ as required.