# 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？

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How to? I will give you tips but the whole solution is a bit long.

The most important thing to remember is if you can rearrange an expression to be the sum of the two squares then you can easily find the minimum. Take $x^2+4*x*y+9y^2=(x+2y)^2+5y^2$, this will have a minimum where $y=0$ and $x+2*y=0$.

So we want to rearrange $$x^2+5y^2+8z^2 = (\alpha x + \beta y + \gamma z)^2 + (\delta x + \epsilon y + \eta z)^2 + \zeta (xy+yz+zx)$$ to some values of these constants. Given how the first two are squares they will never be negative. If we could calculate all the constants and $\zeta$ would be negative then it'd be easy to say the minimal is where the first two squares are zero (or minimal). But there are too many constants so there would be too many solutions. We can try to arbitrarily add some constraints to them -- it wouldn't change what are we looking for.

$$x^2+5y^2+8z^2 = (x + b y + b z)^2 + (cy + dz)^2 + e (xy+yz+zx)$$

This now can be solved for b, c, d and e. Once again note you do not need to consider every possible solution. If you can find one solution where you can easily figure out the minimum of the first two (and $e$ is negative), then you are done.

Also note you will find a possible minimum, don't forget to provide actual $x,y,z$ values.

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Very nice. Looking at the first display, and comparing coefficients of $x^2,y^2,z^2$, we get the equations $\alpha^2+\delta^2=1,\beta^2+\epsilon^2=5,\gamma^2+\eta^2=8$, which have some pretty obvious integer solutions. I suppose we're a bit lucky that everything works out so nicely. –  Gerry Myerson Nov 25 '11 at 9:49

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 $\left\{\frac{3}{2},-\frac{1}{2},-\frac{1}{4}\right\}$.

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There's absolutely no way this is correct. Even if we restrict to the surface $z=0$ then we get $xy=-1$ and then it's quite to trivial tosee $x^2+5y^2 = 5y^2 - y^\left(-2\right)$ has no maximum. –  chx Nov 25 '11 at 9:10
You are correct. All that I have shown is that these two points are the critical points. The surface $xy+yz+zx=-1$ is unbounded so $x^2+5y^2+8z^2$ cannot have a maximum. I will adjust my answer. –  robjohn Nov 25 '11 at 9:18
Also "first variation" and matrixes as elementary? Matrixes maybe but variation?? –  chx Nov 25 '11 at 9:34
This is not using anything more complex than looking at small changes of $x,y,z$. The notation is simpler than, but equivalent to, using partial derivatives. We could also substitute $x=-\frac{1+yz}{y+z}$ into $x^2+5y^2+8z^2$ and take partials, but I thought this method might be of interest. –  robjohn Nov 25 '11 at 10:08

The surface $H: h(x,y,z):=xy + yz + zx +1=0$ is a hyperboloid of one sheet and extends to infinity. On the other hand the level surfaces of the function $f(x,y,z):=x^2+5y^2+8z^2$ are ellipsoids centered at $(0,0,0)$. We have to find the smallest such ellipsoid that meets $H$. Starting with a tiny ellipsoid $E$ and blowing it up we see that when $E$ meets $H$ for the first time at some point $(x_0,y_0,z_0)\in H$ then $E$ and $H$ will actually just touch and so have a common tangent plane $T$ at $(x_0,y_0,z_0)$.

Now $T$ is perpendicular to both gradients $\nabla h(x_0,y_0,z_0)$ and $\nabla f(x_0,y_0,z_0)$; therefore these two gradients are parallel, which means that there is a $\lambda\in{\mathbb R}$ with $\nabla h(x_0,y_0,z_0)=\lambda \nabla f(x_0,y_0,z_0)$. It follows that the point $(x_0,y_0,z_0)\ne(0,0,0)$ satisfies the system

$$y+z =2\lambda x, \quad x+z=10\lambda y,\quad x+y=16\lambda z \qquad(*)$$

for some $\lambda$. This is a homogeneous linear system with matrix

$$\left[\matrix{2\lambda &-1&-1 \cr -1&10\lambda& -1 \cr -1&-1& 16\lambda\cr}\right]\ ,$$

and it will only have a solution $\ne(0,0,0)$ when its determinant is $=0$. This leads to the equation $320\lambda^3-28\lambda-2=0$ which has the three solutions

$$\lambda_1=-{1\over4},\quad \lambda_2={5-\sqrt{65}\over 40},\quad \lambda_3={5+\sqrt{65}\over 40}\ .$$

For each of these $\lambda_i$ we have to solve the system $(*)$. We shall get three vectors $v_i$, each of them determined up to a multiplicative scalar.

This means that the point $(x_0,y_0,z_0)$ we are looking for is a multiple of one of the $v_i$. Each line $t\mapsto t v_i$ intersects the hyperboloid $H$ in two points $\pm(x_i,y_i,z_i)$, so that we obtain altogether 6 candidates for $(x_0,y_0,z_0)$. By comparing the three values $f(x_i,y_i,z_i)$ we finally can determine the actual minimum (it is taken in two points).

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I guess OP hasn't defined "elementary". I imagined it to mean "not using calculus," as in the solution by chx. –  Gerry Myerson Nov 25 '11 at 22:50