Given $a,b,c\ni \mathbb {R_{>0}}$ s.t. $abc\left(a+b+c\right)=1$, find the maximum of $\left(a+b\right)\left(a+c\right)$

Question

Given $a,b,c\ni \mathbb {R_{>0}}$ s.t. $abc\left(a+b+c\right)=1$, find the maximum of $\left(a+b\right)\left(a+c\right)$.

I have been confused by this question for a week, but I can't find any clues of solving this question. Although, I find that $$\left(a+b\right)\left(a+c\right)=bc+a\left(a+b+c\right)$$ and $$abc\left(a+b+c\right)=bc\times a\left(a+b+c\right)=1.$$ I can only say that the minimum is $2$, but not the maximum. If you solve it or you find some clues that may help solving this question, please tell me! Thanks!

Answer 1

From your first equation $$(a+b)(a+c)=a(a+b+c)+bc=\frac{1}{bc}+bc.$$ Now let $b=c=x$. Then by solving $ax^2\left(a+2x\right)=1$ we have that given $x>0$, for $$a=\frac{-x^2+\sqrt{x^4+1}}{x}>0$$ the condition $abc\left(a+b+c\right)=1$ is satisfied. Hence, by letting $x\to +\infty$, it follows that the "maximum" of $\frac{1}{bc}+bc=\frac{1}{x^2}+x^2$ is $+\infty$ .

Answer 2

The maximum does not exist.

Try $c=1$, $b=\frac{-a(a+1)+\sqrt{a^2(a+1)^2+4a}}{2a}$ and $a\rightarrow+\infty$.