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

Solution 1:

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 $$(a+b+c)^2\le 12$$

Solution 2:

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 $$ 3\le a+b+c\le2\sqrt3\tag{8} $$