<s>[INST] Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.

I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
 [/INST] $a=\dfrac{b+c}{bc-1}>0 \to bc>1 \implies $ two of ${a,b,c} >1 $, otherwise ,one will be negative.
WOLG, let $a>1,b>1$. if $c\le 1, \implies (a-1)(b-1)(c-1) \le 0$ so the possible max value will be got when $c>1$.
$x=a-1>0,y=b-1>0,z=c-1>0, x+y+z+3=xyz+x+y+z+xy+yz+xz+1 \to xyz+xy+yz+xz=2$
$xy+yz+xz \ge 3(xyz)^{\frac{2}{3}} .\ \  u=(xyz)^{\frac{1}{3}}>0 \to u^3+3u^2 \le2 \iff u^3+3u^2-2 \le 0 \iff (u+1)(u^2+2u-2) \le 0 \to (u+1+\sqrt{3})(u+1-\sqrt{3})\le 0 \iff u \le \sqrt{3}-1 \implies xyz \le (\sqrt{3}-1)^3 \implies [(a-1)(b-1)(c-1)]_{max}=(\sqrt{3}-1)^3$  
the "=" will hold when $x=y=z=\sqrt{3}-1 \implies a=b=c=\sqrt{3}$
</s>