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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:

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

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

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

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

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

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.

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 $X,Y,Zinmathbb{R}$, 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:
$$begin{array}{ll}text{optimize}&F(X,Y,Z) \text{subject to} & X,Y,Zinmathbb{R}\&G(X,Y,Z)=0,.end{array}$$

We set up the Lagrangian $mathcal{L}(X,Y,Z,Lambda)$ for $X,Y,Z,Lambdainmathbb{R}$ by
$$mathcal{L}(X,Y,Z,Lambda):=F(X,Y,Z)-3,Lambda,G(X,Y,Z),.$$
If $(x,y,z)inmathbb{R}^3$ is such that $(X,Y,Z):=(x,y,z)$ is a solution to this optimization problem, then there exists $lambdainmathbb{R}$ for which
$$frac{partial mathcal{L}}{partial V}(x,y,z,lambda)=0$$
for all variables $Vin{X,Y,Z,Lambda}$. 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
$$begin{align}f&=x(x^2-yz)+y(y^2-zx)+z(z^2-xy)=x(lambda,a)+y(lambda,b)+z(lambda,c)\&=lambda,(ax+by+cz)=lambdacdot 1=lambda,,end{align}$$ 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 $f=0$.

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 $a+b+c=0$.

Case II: $a+b+cneq 0$ but $f=0$. By adding (1), (2), and (3) together, we conclude, as in Case I, that $x=y=z$. Ergo,
$$1=ax+by+cz=ax+bx+cx=(a+b+c)xtext{ implies }x=y=z=frac{1}{a+b+c},.$$
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.

Case III: $a+b+cneq 0$ and $fneq 0$. 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
$$begin{align}&(a^3+b^3+c^3-3abc)f\&phantom{aaa}=frac{3big((a+b+c)(a^2+b^2+c^2)f-(a+b+c)kbig)-(a+b+c)big((a+b+c)^2f-3kbig)}{2}\&phantom{aaa}=frac{3cdot 1-(a+b+c)cdotleft(frac{1}{a+b+c}right)}{2}=1,.end{align}$$
Consequently, $a^3+b^3+c^3-3abcneq 0$, 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,
$$begin{align}(f,x,y,z)&=Biggl(frac{1}{a^3+b^3+c^3-3abc},frac{a^2-bc}{a^3+b^3+c^3-3abc}\&phantom{aaaaa},frac{b^2-ca}{a^3+b^3+c^3-3abc},frac{c^2-ab}{a^3+b^3+c^3-3abc}Biggr),.end{align}tag{6}$$

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 $X,Y,Z,Lambdainmathbb{R}$, let
$$mathcal{H}(X,Y,Z,Lambda):=begin{bmatrix}frac{partial^2mathcal{L}}{partial Lambda^2}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partialLambda,partial X}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partialLambda,partial Y}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partialLambda,partial Z}(X,Y,Z,Lambda)\
frac{partial^2mathcal{L}}{partial X,partial Lambda}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial X^2}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial X,partial Y}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial X,partial Z}(X,Y,Z,Lambda) \
frac{partial^2mathcal{L}}{partial Y,partial Lambda}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial Y,partial X}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial Y^2}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial Y,partial Z}(X,Y,Z,Lambda)
\
frac{partial^2mathcal{L}}{partial Z,partial Lambda}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial Z,partial X}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial Z,partial Y}(X,Y,Z,Lambda) & frac{partial^2mathcal{L}}{partial Z^2}(X,Y,Z,Lambda)
end{bmatrix},,$$
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 $X,Y,Z,Lambdainmathbb{R}$. We shall write $H:=mathcal{H}(x,y,z,lambda)=mathcal{H}(x,y,z,f)$. Recall that $a+b+cneq 0$ for $(f,x,y,z)$ to exist. For an $n$-by-$n$ matrix $X$ and $min{1,2,ldots,n}$, 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 $aneq 0$. 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 $a+bneq 2c$ and so, for any $tinmathbb{R}$, there exists $(u_t,v_t,w_t)inmathbb{R}^3$ 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, $$F(X,Y,Z)geq 0text{ when }a>0,,$$ 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, $$F(X,Y,Z)leq 0text{ when }a<0,,$$ and so $f=0$ is the global maximum value of $F$ on the solution set of $G=0$, where the globally maximizing point is $(x,y,z)=left(dfrac1{3a},dfrac1{3a},dfrac1{3a}right)$.

Without Lagrange Multipliers.

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

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

or equivalently

$$
minmax f(lambda,mu) = frac{1+lambda^3+mu^3-3lambdamu}{(a+lambda b+mu c)^3}
$$

whose stationary points are solved by

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

giving the points

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