# Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle

Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.

The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?

• @lhf By rearrangement. Jan 21, 2016 at 14:47

We have to use the word "triangle" at some point. $$a^2+b^2+c^2 \le a(b+c)+b(a+c)+c(a+b)=2(ab+ac+bc)=6$$ hence $$(a+b+c)^2\le 12$$

• Shouldn't it be <and not ≤ since it is a triangle? Jan 21, 2016 at 15:07
• @Puzzled417 Indeed, if we assume positive side lengths and strict triangle inequality, all goes through with $<$, ultimately showing $a+b+c<2\sqrt 3$. Jan 21, 2016 at 20:44

While working on a non-variational approach, I present a variational apporach.

By the given condition, $ab+bc+ca=3$, we get $$(b+c)\,\delta a+(c+a)\,\delta b+(a+b)\,\delta c=0\tag{1}$$ For interior critical points maximizing $a+b+c$, we want $$\delta a+\delta b+\delta c=0\tag{2}$$ To get $(2)$ for all variations that satisfy $(1)$, we get $b+c=c+a=a+b$, which means $a=b=c=1$. That is $$a+b+c=3\tag{3}$$ The edge critical points will come when either $a=b+c$ or $b=c+a$ or $c=a+b$. Without loss of generality, assume $a=b+c$. $(1)$ becomes $$(2b+3c)\,\delta b+(3b+2c)\,\delta c=0\tag{4}$$ and $(2)$ becomes $$2\delta b+2\delta c=0\tag{5}$$ To get $(5)$ for all variations that satisfy $(4)$, we get $2b+3c=3b+2c$, which means $b=c=\sqrt{\frac35}$. That is $$a+b+c=4\sqrt{\frac35}\tag{6}$$ The corner critical points come from $a=0$ or $b=0$ or $c=0$. Without loss of generality, assume $c=0$, which means $a=b=\sqrt3$. That is $$a+b+c=2\sqrt3\tag{7}$$ Combining $(3)$, $(6)$, and $(7)$ gives $$3\le a+b+c\le2\sqrt3\tag{8}$$

Why is $a^2+b^2+c^2\geq 3$? I think you need to apply somewhere that you are dealing with a triangle. Maybe something like this?

If you sum up the following inequalities,

$$ab+bc=b(a+c)> b^2$$ $$ab+ac=a(b+c)> a^2$$ $$ac+bc=c(a+b)> c^2$$

you obtain $a^2+b^2+c^2\leq 6$ which by your argument gives you $(a+b+c)^2<12.$ This implies $a+b+c\leq 2< 3.$ However, the left-hand side of the inequality follows from Cauchy-Schwarz inequality for triples $(a,b,c)$ and $(b,c,a)$ and $$ab+bc+ca\leq \sqrt{a^2+b^2+c^2} \cdot \sqrt{b^2+c^2+a^2}=a^2+b^2+c^2.$$ This finally implies $a^2+b^2+c^2\geq 3$ as you assumed before.

• We have $a^2+b^2+c^2\geq ab+bc+ca\geq 3$, by the Rearrangement Inequality
– user304329
Jan 21, 2016 at 14:49
• @vrugtehagel I understand now. It is sometime now I heard about this inequality. I remember it now. Jan 21, 2016 at 14:51
• @Marko Shouldn't it be $>$ and not $≥$ since it is a triangle? Jan 21, 2016 at 15:07
• Ofcourse, I wrote $\leq$ by default. Jan 21, 2016 at 15:09

For the other side, we note that \begin{align*} (a+b+c)^2 &\ge 3 (ab + bc + ca)\\ &= 9. \end{align*}