<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] 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...
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