Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers

Question

If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but second is not an extreme value. Comment on particular cases when (i) ${a+b+c=0}$, (ii) ${a=b=c}$.

My Attempt:

$${F=f+\lambda \phi =x^3 + y^3 + z^3 - 3xyz + 3\lambda(ax + by + cz-1)}$$ $${\frac13F_x = x^2-yz+\lambda a = 0}{\text{ ...(1)}}$$ $${\frac13F_y = y^2-xz+\lambda b = 0}{\text{ ...(2)}}$$ $${\frac13F_z = z^2-xy+\lambda c = 0{\text{ ...(3)}}}$$ $${(1)x+(2)y+(3)z \implies f+\lambda (1) = 0 \implies \lambda = -f}$$

$${(1)+(2)+(3) \implies}{x^2+y^2+z^2-xy-yz-zx=(a+b+c)f}$$

$${\implies f/(x+y+z)=(a+b+c)f}\,,$$ then ${f=0}$ or ${x+y+z=1/(a+b+c)}$.

Also, ${(1)-(2) \implies x^2-y^2-z(x-y)=f(a-b) \implies \frac{x-y}{a-b}=f\frac{a+b+c}{x+y+z}}$

Similarly ${(2)-(3)}$ and ${(3)-(1)}$, then we get ${\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}=f\frac{a+b+c}{x+y+z}}$

I don't know how to proceed further. I couldn't get the other stationary value of ${f}$. I need help in proceeding further to calculate stationary points and/or stationary values, if at all, my method is correct.

Answer 1

Here I shall find all critical points without verifying what kind of critical points they are (which seems to be your question). My notations are similar to yours, but a bit different, so I shall work from scratch but my answer borrows a lot of your ideas (great attempt, by the way). However, I do not expect that any optimizing point will be a global one. Therefore, be careful and do not assume that any optimizing point will yield a global optimum.

For $X,Y,Z\in\mathbb{R}$, let $$\begin{align}F(X,Y,Z)&:=X^3+Y^3+Z^3-3XYZ\\&=(X+Y+Z)(X^2+Y^2+Z^2-YZ-ZX-XY)\end{align}$$ and $$G(X,Y,Z):=aX+bY+cZ-1\,,$$ where $a$, $b$, and $c$ are fixed real numbers. The task is as follows: $$\begin{array}{ll}\text{optimize}&F(X,Y,Z) \\\text{subject to} & X,Y,Z\in\mathbb{R}\\&G(X,Y,Z)=0\,.\end{array}$$

We set up the Lagrangian $\mathcal{L}(X,Y,Z,\Lambda)$ for $X,Y,Z,\Lambda\in\mathbb{R}$ by $$\mathcal{L}(X,Y,Z,\Lambda):=F(X,Y,Z)-3\,\Lambda\,G(X,Y,Z)\,.$$ If $(x,y,z)\in\mathbb{R}^3$ is such that $(X,Y,Z):=(x,y,z)$ is a solution to this optimization problem, then there exists $\lambda\in\mathbb{R}$ for which $$\frac{\partial \mathcal{L}}{\partial V}(x,y,z,\lambda)=0$$ for all variables $V\in\{X,Y,Z,\Lambda\}$. That is, we have the following equations: $$x^2-yz=\lambda\,a\,,\tag{1}$$ $$y^2-zx=\lambda\,b\,,\tag{2}$$ $$z^2-xy=\lambda\,c\,,\tag{3}$$ and $$ax+by+cz=1\,.\tag{4}$$

We let $f:=F(x,y,z)$. Adding $x$ times (1), $y$ times (2), and $z$ times (3) yields $$\begin{align}f&=x(x^2-yz)+y(y^2-zx)+z(z^2-xy)=x(\lambda\,a)+y(\lambda\,b)+z(\lambda\,c)\\&=\lambda\,(ax+by+cz)=\lambda\cdot 1=\lambda\,,\end{align}$$ due to (4). By adding (1), (2), and (3), we obtain $$\begin{align} f&=(x+y+z)(x^2+y^2+z^2-yz-zx-xy) \\&=(x+y+z)\big((x^2-yz)+(y^2-zx)+(z^2-xy)\big) \\&=(x+y+z)\big(\lambda\,a+\lambda\,b+\lambda\,c)=(x+y+z)(a+b+c)\lambda\\&=(x+y+z)(a+b+c)f\,. \end{align}$$ This means $(x+y+z)(a+b+c)=1$ or $f=0$.

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Case I: $a+b+c=0$. Then, $f=0$ must hold (whence $\lambda=f=0$). By adding (1), (2), and (3) together, we obtain $$\frac{(y-z)^2+(z-x)^2+(x-y)^2}{2}=(x^2-yz)+(y^2-zx)+(z^2-xy)=0\,.$$ Thus, $x=y=z$ must be the case. Ergo, $$1=ax+by+cz=ax+bx+cx=(a+b+c)x=0\,,$$ which is a contradiction. Consequently, there does not exist a critical point when $a+b+c=0$.

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Case II: $a+b+c\neq 0$ but $f=0$. By adding (1), (2), and (3) together, we conclude, as in Case I, that $x=y=z$. Ergo, $$1=ax+by+cz=ax+bx+cx=(a+b+c)x\text{ implies }x=y=z=\frac{1}{a+b+c}\,.$$ This yields $$(f,x,y,z)=\left(0,\frac1{a+b+c},\frac1{a+b+c},\frac1{a+b+c}\right)\,.\tag{5}$$ We shall see later that, when $a=b=c$, then this is the only case that yields a critical point.

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Case III: $a+b+c\neq 0$ and $f\neq 0$. Then, we must have $x+y+z=\dfrac1{a+b+c}$. Subtracting (2) from (1) gives us $(x-y)(x+y+z)=\lambda(a-b)=f(a-b)$, so that $$x-y=(a+b+c)f(a-b)\,.$$ Similarly, $$y-z=(a+b+c)f(b-c)$$ and $$z-x=(a+b+c)f(c-a)\,.$$ Set $k:=(a+b+c)fa-x$. Then, we get $$x=(a+b+c)fa-k\,,\,\,y=(a+b+c)fb-k\,,\text{ and }z=(a+b+c)fc-k\,.$$ As $x+y+z=\dfrac1{a+b+c}$, we obtain $$(a+b+c)^2f-3k=\frac{1}{a+b+c}\,.$$ Because $ax+by+cz=1$, we must have $$(a+b+c)(a^2+b^2+c^2)f-(a+b+c)k=1\,.$$ This shows that $$\begin{align}&(a^3+b^3+c^3-3abc)f\\&\phantom{aaa}=\frac{3\big((a+b+c)(a^2+b^2+c^2)f-(a+b+c)k\big)-(a+b+c)\big((a+b+c)^2f-3k\big)}{2}\\&\phantom{aaa}=\frac{3\cdot 1-(a+b+c)\cdot\left(\frac{1}{a+b+c}\right)}{2}=1\,.\end{align}$$ Consequently, $a^3+b^3+c^3-3abc\neq 0$, implying that $a$, $b$, and $c$ are not all equal, and so $$f=\frac{1}{a^3+b^3+c^3-3abc}\,,\text{ which leads to }k=\frac{bc+ca+ab}{a^3+b^3+c^3-3abc}\,.$$ Thence, $$\begin{align}(f,x,y,z)&=\Biggl(\frac{1}{a^3+b^3+c^3-3abc},\frac{a^2-bc}{a^3+b^3+c^3-3abc}\\&\phantom{aaaaa},\frac{b^2-ca}{a^3+b^3+c^3-3abc},\frac{c^2-ab}{a^3+b^3+c^3-3abc}\Biggr)\,.\end{align}\tag{6}$$

Answer 2

This is a proof whether a critical point found in my other answer is a local or global minimum, a local or global maximum, or neither. I have to write a separate answer because MathJax is being a pain when I write a long answer. I refer to notations in my other answer, and this answer is based on here.

For $X,Y,Z,\Lambda\in\mathbb{R}$, let $$\mathcal{H}(X,Y,Z,\Lambda):=\begin{bmatrix}\frac{\partial^2\mathcal{L}}{\partial \Lambda^2}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial\Lambda\,\partial X}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial\Lambda\,\partial Y}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial\Lambda\,\partial Z}(X,Y,Z,\Lambda)\\ \frac{\partial^2\mathcal{L}}{\partial X\,\partial \Lambda}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial X^2}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial X\,\partial Y}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial X\,\partial Z}(X,Y,Z,\Lambda) \\ \frac{\partial^2\mathcal{L}}{\partial Y\,\partial \Lambda}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial Y\,\partial X}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial Y^2}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial Y\,\partial Z}(X,Y,Z,\Lambda) \\ \frac{\partial^2\mathcal{L}}{\partial Z\,\partial \Lambda}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial Z\,\partial X}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial Z\,\partial Y}(X,Y,Z,\Lambda) & \frac{\partial^2\mathcal{L}}{\partial Z^2}(X,Y,Z,\Lambda) \end{bmatrix}\,,$$ which is the bordered Hessian matrix of the Lagrangian $\mathcal{L}(X,Y,Z,\Lambda)$. That is, $$\mathcal{H}(X,Y,Z,\Lambda)=3\,\begin{bmatrix} 0 & -a & -b & -c \\ -a & 2X & -Z & -Y \\ -b & -Z & 2Y & -X \\ -c & -Y & -X & 2Z \end{bmatrix}\,,$$ for all $X,Y,Z,\Lambda\in\mathbb{R}$. We shall write $H:=\mathcal{H}(x,y,z,\lambda)=\mathcal{H}(x,y,z,f)$. Recall that $a+b+c\neq 0$ for $(f,x,y,z)$ to exist. For an $n$-by-$n$ matrix $X$ and $m\in\{1,2,\ldots,n\}$, write $X_m$ for the principal minor of $X$ which is the $m$-by-$m$ matrix consisting of the truncated first $m$ rows and $m$ columns of $X$.

In the case $(f,x,y,z)$ is given by (5) in my previous answer, then $$H=\frac{3}{a+b+c}\,\begin{bmatrix} 0 & -a(a+b+c) & -b(a+b+c) & -c(a+b+c)\\ -a(a+b+c) & 2 & -1 & -1 \\ -b(a+b+c) & -1 & 2 & -1 \\ -c(a+b+c) & -1 & -1 &2 \end{bmatrix}\,.$$ Then, $$\det(H_3)=-\dfrac{54(a^2+ab+b^2)}{a+b+c}\text{ and }\det(H_4)=\det(H)=-243\,.$$ Without loss of generality, we may permute $a$, $b$, and $c$ so that $a\neq 0$. On one hand, if $a+b+c>0$, then $\det(H_3)$ and $\det(H_4)$ are both negative, so $f$ is a local minimum value corresponding to locally minimizing point $(x,y,z)$. On the other hand, if $a+b+c<0$, then $\det(H_3)>0$ and $\det(H_4)<0$, so $f$ is a local maximum value corresponding to the locally maximizing point $(x,y,z)$.

In the case that $(f,x,y,z)$ is given by (6) in my previous answer (whence $a$, $b$, and $c$ are not all equal), we get $$H=\frac{3}{s}\,\begin{bmatrix} 0 & -sa & -sb & -sc \\ -sa & 2(a^2-bc) & -(c^2-ab) & -(b^2-ca) \\ -sb & -(c^2-ab) & 2(b^2-ca) & -(a^2-bc) \\ -sc & -(b^2-ca) & -(a^2-bc) & 2(c^2-ab) \end{bmatrix}\,,$$ where $s:=a^3+b^3+c^3-3abc$. Then, $$\det(H_3)=-\dfrac{54(a^2-bc)(b^2-ca)}{s}\text{ and }\det(H_4)=\det(H)=81>0\,.$$ Hence, $f$ is not a local optimum value, as $(x,y,z)$ is a saddle point.

Note that $f$ is never a global optimum if $a$, $b$, and $c$ are not all equal. This is because we may assume without loss of generality that $a+b\neq 2c$ and so, for any $t\in\mathbb{R}$, there exists $(u_t,v_t,w_t)\in\mathbb{R}^3$ such that $au_t+bv_t+cw_t=1$, $u_t+v_t+w_t=t$, and $u_t-v_t=t$. Therefore, $$F(u_t,v_t,w_t)=(u_t+v_t+w_t)\left(\frac{(v_t-w_t)^2+(w_t-u_t)^2+(u_t-v_t)^2}{2}\right)$$ implies that $$F(u_t,v_t,w_t)\geq \frac{t^3}{2}\text{ if }t>0$$ and $$F(u_t,v_t,w_t)\leq \frac{t^3}{2}\text{ if }t<0\,.$$ Ergo, $F$ is unbounded both from above and below on the solution set of $G=0$.

However, if $a=b=c$, then $G(X,Y,Z)=0$ implies $X+Y+Z=\dfrac1a$. Consequently, $$F(X,Y,Z)=\dfrac1a\,\left(\frac{(Y-Z)^2+(Z-X)^2+(X-Y)^2}{2}\right)\,.$$ Thus, $$F(X,Y,Z)\geq 0\text{ when }a>0\,,$$ implying that $f=0$ is the global minimum value on the solution set of $G=0$, and the globally minimizing point is $(x,y,z)=\left(\dfrac1{3a},\dfrac1{3a},\dfrac1{3a}\right)$. Finally, $$F(X,Y,Z)\leq 0\text{ when }a<0\,,$$ and so $f=0$ is the global maximum value of $F$ on the solution set of $G=0$, where the globally maximizing point is $(x,y,z)=\left(\dfrac1{3a},\dfrac1{3a},\dfrac1{3a}\right)$.

Answer 3

Without Lagrange Multipliers.

Making the change of variables $y = \lambda x, z = \mu x$ and substituting we get

$$ \min\max x^3(1+\lambda^3+\mu^3-3\lambda\mu)\ \ \mbox{s. t. }\ \ x(a+\lambda b+\mu c) = 1 $$

or equivalently

$$ \min\max f(\lambda,\mu) = \frac{1+\lambda^3+\mu^3-3\lambda\mu}{(a+\lambda b+\mu c)^3} $$

whose stationary points are solved by

$$ \left\{ \begin{array}{rcl} \left(\lambda ^2-\mu \right) (a+c \mu )-b \left(\mu ^3-2 \lambda \mu +1\right)& = & 0 \\ c \left(\lambda ^3-2 \mu \lambda +1\right)+(a+b \lambda ) \left(\lambda -\mu ^2\right) & = & 0 \\ \end{array} \right. $$

giving the points

$$ \begin{array}{ccc} \lambda & \mu & f(\lambda,\mu)\\ 1 & 1 & 0 \\ \frac{b^2-a c}{a^2-b c} & \frac{c^2-a b}{a^2-b c} & \frac{1}{a^3+b^3+c^3-3 a b c} \\ \end{array} $$