Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$.

Question

If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.

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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).

Answer 1

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!

Answer 2

$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 $x=y=z=\sqrt{3}-1 \implies a=b=c=\sqrt{3}$

Answer 3

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...

Answer 4

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!