find cube roots of $x^3+y^3+z^3=3xyz+1$.

Question

Let $x$,$y$, $z$ be real numbers such that $x^3+y^3+z^3 = 3xyz + 1$.

What is the smallest possible value of $x^2+y^2+z^2$?

Answer 1

Lagrange Approach

At any critical point of $$ x^2+y^2+z^2\tag1 $$ under the condition $$ x^3+y^3+z^3=3xyz+1\tag2 $$ Lagrange says we need a constant $\lambda$ so that $$ \begin{align} x(x-\lambda)=yz\tag3\\ y(y-\lambda)=zx\tag4\\ z(z-\lambda)=xy\tag5 \end{align} $$ Multiplying $(3)$ by $x$, $(4)$ by $y$, and $(5)$ by $z$ yields $$ \begin{align} x^3-\lambda x^2=xyz\tag{6}\\ y^3-\lambda y^2=xyz\tag{7}\\ z^3-\lambda z^2=xyz\tag{8} \end{align} $$ Summing $(6)$, $(7)$, and $(8)$ says $$ 1=x^3+y^3+z^3-3xyz=\lambda\left(x^2+y^2+z^2\right)\tag9 $$ Note that $(9)$ says that $\lambda\gt0$.

Case $\bf{1}$: One of $\boldsymbol{x,y,z}$ is $\bf{0}$

Suppose one of $x,y,z$ is $0$. Without loss of generality, say $x=0$. $(3)$ then implies that one or both of $y,z$ must also be $0$. Without loss of generality, say $y=0$. $(2)$ then implies that $z=1$. Thus, $x^2+y^2+z^2=1$.

Case $\bf{2}$: None of $\boldsymbol{x,y,z}$ is $\bf{0}$

Suppose none of $x,y,z$ is $0$. $(3)$, $(4)$, and $(5)$ imply that $$ \begin{align} (x-\lambda)(y-\lambda)=z^2\tag{10}\\ (y-\lambda)(z-\lambda)=x^2\tag{11}\\ (z-\lambda)(x-\lambda)=y^2\tag{12} \end{align} $$ Combining $(3)$ and $(11)$ gives $$ \begin{align} yz &=x^2-\lambda x\\ &=(y-\lambda)(z-\lambda)-\lambda x\\ 0 &=\lambda^2-\lambda(x+y+z)\tag{13} \end{align} $$ Therefore, $$ \lambda=x+y+z\tag{14} $$ Adding $(3)$, $(4)$, and $(5)$ and applying $(14)$ yields $$ xy+yz+zx=0\tag{15} $$ which means, via $(9)$, $(14)$, and $(15)$, $$ \overbrace{x^2+y^2+z^2}^{1/\lambda}=\overbrace{(x+y+z)^2}^{\lambda^2}\tag{16} $$ Thus, $\lambda=1$, and $x^2+y^2+z^2\ge1$.

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$\boldsymbol{x^2+y^2+z^2}$ Can be Arbitrarily Large

Note that we can make $x^2+y^2+z^2$ as large as we wish: let $y=-x$ and $x$ be arbitrarily large. Then $x^3+y^3+z^3-3xyz-1=z^3+3x^2z-1=0$ will always be satisfied by some real $z$. Then $x^2+y^2+z^2\ge2x^2$.

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Factorization Approach

The Lagrange approach says that the minimum is $1$, so there is a good chance that the difference $$ \left(x^2+y^2+z^2\right)^3-\left(x^3+y^3+z^3-3xyz\right)^2 $$ might vanish at the critical points. The Lagrange approach also says that, at the critical points, $xy+yz+zx=0$, so it might be a factor. In fact, its square is a factor. The remaining factor is $$ 3x^2+3y^2+3z^2-2xy-2yz-2zx=(x+y-z)^2+(y+z-x)^2+(z+x-y)^2 $$ That is, $$ \begin{align} &\left(x^2+y^2+z^2\right)^3-\overbrace{\left(x^3+y^3+z^3-3xyz\right)^2}^{\large1}\\ &=(xy+yz+zy)^2\left[(x+y-z)^2+(y+z-x)^2+(z+x-y)^2\right]\\ &\ge0 \end{align} $$ Since $(0,0,1)$ satisfies $(2)$, the minimum of $(1)$ is $1$.