<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] Go for Lagrange Multipliers!  Let $f(a,b,c)=(a-1)(b-1)(c-1)$, and $g(a,b,c)=abc-a-b-c$. Then you are trying to maximize $f(a,b,c)$ subject to $g(a,b,c)=0$.
Now, we have
$$
\nabla f(a,b,c)=\bigl\langle(b-1)(c-1),(a-1)(c-1),(a-1)(b-1)\bigr\rangle
$$
and
$$
\nabla g(a,b,c)=\langle bc-1, ac-1, ab-1\rangle.
$$
So, you want to find all tuples $(a,b,c,\lambda)$ such that
$$
\begin{align}
\tag{1} (b-1)(c-1)&=\lambda(bc-1)\\
\tag{2} (a-1)(c-1)&=\lambda(ac-1)\\
\tag{3} (a-1)(b-1)&=\lambda(ab-1)\\
\tag{4} abc-a-b-c&=0
\end{align}
$$
Go through and solve this system of equations.  There are only 3 solutions!
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