Find the minimum of $x^{2}+5y^{2}+8z^{2}$ if $xy+yz+zx=-1$ 
If  $xy+yz+zx=-1$，find the minimum of  $x^{2}+5y^{2}+8z^{2}$.

How to solve it use Elementary mathematics methods？

Taking the first variation of $xy+yz+zx=-1$, we get
$$
(y+z)\delta x+(z+x)\delta y+(x+y)\delta z=0\tag{1}
$$
Setting the first variation of $x^2+5y^2+8z^2$ to $0$, we get
$$
2x\delta x+10y\delta y+16z\delta z=0\tag{2}
$$
At an extreme $\{x,y,z\}$,  any $\{\delta x,\delta y,\delta z\}$ that satisfies $(1)$, must also satsify $(2)$. Orthogonality dictates that
$$
2\lambda x=y+z\qquad 10\lambda y=z+x\qquad 16\lambda z=x+y\tag{3}
$$
Converting $(3)$ to a matrix yields:
$$
\begin{bmatrix}-\lambda&1/2&1/2\\1/10&-\lambda&1/10\\1/16&1/16&-\lambda\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=0\tag{4}
$$
Thus, $\begin{bmatrix}x&y&z\end{bmatrix}^T$ will be an eigenvector of
$$
\begin{bmatrix}0&1/2&1/2\\1/10&0&1/10\\1/16&1/16&0\end{bmatrix}\tag{5}
$$
which has eigenvectors
$$
v_1=\begin{bmatrix}\frac{7+\sqrt{65}}{4}\\\frac{5+3\sqrt{65}}{20}\\1\end{bmatrix}\qquad v_2=\begin{bmatrix}\frac{7-\sqrt{65}}{4}\\\frac{5-3\sqrt{65}}{20}\\1\end{bmatrix}\qquad v_3=\begin{bmatrix}-6\\2\\1\end{bmatrix}\tag{6}
$$
Computing $xy+yz+zx$ for these vectors, we get
$$
v_1\mapsto\frac{195+29\sqrt{65}}{40}\qquad v_2\mapsto\frac{-416}{195+29\sqrt{65}} \qquad v_3\mapsto-16\tag{7}
$$
Since $xy+yz+zx\ge0$ for $v_1$, there is no multiple of $v_1$ so that $xy+yz+zx=-1$. That leaves $v_2$ and $v_3$ as the extrema.
If we scale $v_2$ so that $xy+yz+zx=-1$, we get $x^2+5y^2+8z^2=5+\sqrt{65}$.
If we scale $v_3$ so that $xy+yz+zx=-1$, we get $x^2+5y^2+8z^2=4$.
Thus, under the constraint that $xy+yz+zx=-1$, $x^2+5y^2+8z^2$ has a minimum of $4$ at $\{x,y,z\}=\left\{-\frac{3}{2},\frac{1}{2},\frac{1}{4}\right\}$ and $\left\{\frac{3}{2},-\frac{1}{2},-\frac{1}{4}\right\}$.
