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).
 A: 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!
A: $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 $x=y=z=\sqrt{3}-1 \implies a=b=c=\sqrt{3}$
A: Hints: 
Lagrange multipliers. Define
$$H(x,y,z,\lambda):=(x-1)(y-1)(z-1)+\lambda(x+y+z-xyz)$$
$$\begin{align*}H'_x&=(y-1)(z-1)+\lambda(1-yz)=0\iff \lambda=\frac{(y-1)(z-1)}{yz-1}\\
H'_y&=(x-1)(z-1)+\lambda(1-xz)=0\iff \lambda=\frac{(x-1)(z-1)}{xz-1}\\
H'_z&=(y-1)(x-1)+\lambda(1-yx)=0\iff \lambda=\frac{(y-1)(x-1)}{yx-1}\\
H'_\lambda&=x+y+z-xyz=0\end{align*}$$
From the first three equations we get (using symmetry)
$$\frac{x-1}{xy-1}=\frac{z-1}{yz-1}\implies (x-z)(1-y)=0=(x-y)(1-z)=(y-z)(1-x)$$
Check that $\,x=y=z=1\,$ is not a good option...
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$ and we need to find a maximum of
$$abc-ab-ac-bc+a+b+c-1$$ or
$$6u-3v^2-1,$$
which is linear function of $v^2$, 
which says that it's enough to find a maximum for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$.
Hence, $c=\frac{2a}{a^2-1}$, where $a>1$ and we need to find $\max\limits_{a>1}f$, where
$$f(a)=(a-1)^2\left(\frac{2a}{a^2-1}-1\right)^2,$$ 
which gives that 
$$\max\limits_{a>1}f=-10+6\sqrt3.$$
Done!
