2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Don't we need to assume that $a \geq b$ for $a \leq b+c$ from $|a-b| \leq c$ and that $b > a$ for $b \leq a+c$? But then they can't simultaneously be true as you can't assume both $a \geq b$ and $b > a$ right? And we require that they are simultaneously true. Also, I am still unsure as to what you mean by taking the negative square root reverses order. Does this mean that we have both $|a-b| \leq|c|$ and $-|a-b| \geq|c|$ or $-|a-b| \geq-|c|$? $\endgroup$ – user244685 Nov 22 '19 at 4:58
  • $\begingroup$ $|a-b|\leq c$ if and only if both $a-b\leq c$ and $b-a \leq c$ hold. No assumptions are made on $a,b,c$ other than that they are real numbers. By reversing order I mean $-|a-b|\geq -|c|$ $\endgroup$ – lc2r43 Nov 22 '19 at 5:03
  • $\begingroup$ But $|a-b| = a-b$ if $a-b\geq 0 \implies a \geq b$ and $|a-b| = b-a$ if $a-b<0 \implies b>a$ by the properties of absolute value. Do both $a-b \leq c \text { and } b-a \leq c \text { hold }$ by some other rule? Also, by your second remark, $-|a-b| \geq-|c|$ can be rearranged to the original $|a-b| \leq|c|$ so does this mean taking the square root of an inequality does not lead to two 'sets' of solutions as it does for equalities? $\endgroup$ – user244685 Nov 22 '19 at 5:11
  • 1
    $\begingroup$ We always have $a-b\leq |a-b|$ and $b-a \leq |a-b|$, so both $a-b\leq c$ and $b-a\leq c$ hold regardless of whether $a\geq b$ or not. As to the second question, you are correct. For a concrete example, note that I can say $4<9$ so $2<3$, but I cannot also say that $-2<-3$. $\endgroup$ – lc2r43 Nov 22 '19 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.