Minimize $xy$ over $x^2+y^2+z^2=7$ and $xy+xz+yz=4$.

Question

Let $x,$ $y,$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 7$ and $xy + xz + yz = 4.$ Find the smallest possible value of $xy.$

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I used Cauchy to get $$(x^2+y^2+z^2)(1^2+1^2+1^2)\geq (x+y+z)^2$$ and $$(x^2+y^2+z^2)(y^2+z^2+x^2)\geq(xy+xz+yz)^2,$$ but this doesn't do much as both of the inequalities already work. :( Any guidance would be appreciated!!

Thanks in advance!!!

P.S. I've been given that this can be solved by Lagrange multipliers, but I haven't learned it yet. Hopefully, someone can give me hints on a method that doesn't involve Lagrange multipliers. :)

Answer 1

I would start with noticing that

\begin{align*} (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) = 15 \end{align*}

Therefore $x + y + z = \pm\sqrt{15} =: k$. Better saying, $z = k - x - y$.

Then we arrive at the relation

\begin{align*} xy + xz + yz & = xy + z(x + y)\\\\ & = xy + (k - (x + y))(x + y)\\\\ & = xy + k(x + y) - (x + y)^{2} = 4\\\\ & \Rightarrow xy = (x + y)^{2} - k(x + y) + 4 \end{align*}

Let us make the change of variable $t = x + y$.

Then we are interested in the minimum value of the RHS above, which is given by $1/4$ as follows: \begin{align*} t^{2} - kt + 4 & = \left(t^{2} - kt + \frac{k^{2}}{4}\right) + 4 - \frac{k^{2}}{4}\\\\ & = \left(t - \frac{k}{2}\right)^{2} + 4 - \frac{15}{4} \geq \frac{1}{4} \end{align*}

Hence the minimum value of $xy$ satisfying the proposed constraints equals $1/4$.

Hopefully this helps !

EDIT

As an answer to the (well posed) comment of @RobPratt, we must show such value is attained.

Indeed, this is the case as the following system of equations show: \begin{align*} \begin{cases} xy = \dfrac{1}{4}\\\\ x + y = \dfrac{\sqrt{15}}{2} \end{cases} & \Longleftrightarrow \begin{cases} x = \dfrac{\sqrt{15} \mp \sqrt{11}}{4}\\\\ y = \dfrac{\sqrt{15} \pm \sqrt{11}}{4} \end{cases} \end{align*}

as well as the following system of equations:

\begin{align*} \begin{cases} xy = \dfrac{1}{4}\\\\ x + y = -\dfrac{\sqrt{15}}{2} \end{cases} & \Longleftrightarrow \begin{cases} x = -\dfrac{\sqrt{15} \mp \sqrt{11}}{4}\\\\ y = -\dfrac{\sqrt{15} \pm \sqrt{11}}{4} \end{cases} \end{align*}

Answer 2

$x^2+y^2+z^2 = 7$ and $xy +xz + yz = 4$ implies $x+y+z = \pm\sqrt{15}$ Now using the second equality that is given in the question, we get $xy =4 -(x+y)z = 4 - (\pm\sqrt{15} - z)z$

• Case 1: When $x+y+z=\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = \frac{\sqrt{15}}{2}$. So, minimum value of $xy$ is $4 - (\sqrt{15} - z)z= 4-\frac{15}{4}=\frac{1}{4}$.
• Case 2: When $x+y+z=-\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = -\frac{\sqrt{15}}{2}$. So, minimum value of $xy$ is $4 - (-\sqrt{15} - z)z= 4-\frac{15}{4}=\frac{1}{4}$.

So the minimum value of $xy$ is $\frac{1}{4}$.