IMO Optimization Contest Problem: Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.

IMO Optimization Contest Problem:

Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.

My solution:

From the well-known identity

$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$

and the given values for $x^2+y^2+z^2=7$ and $xy+xz+yz=4$, we get:

$(x+y+z)^2=7+2(4)$

$(x+y+z)^2=15$

$x+y+z=\pm\sqrt{15}$

We're asked to look for the minimum $xy$. Therefore, it makes sense to have the plan to get rid of the variable $z$.

Thus we rearrange the equation above and make $z$ the subject:

$\displaystyle \color{yellow}\bbox[5px,purple]{z=\pm\sqrt{15}-(x+y)}$

We then substitute it into the given equation $xy+xz+yz=4$, we have:

$xy+xz+yz=4$

$xy+z(x+y)=4$

$z(x+y)=4-xy$

$z(x+y)=4-xy$

$\displaystyle \color{yellow}\bbox[5px,purple]{(\pm\sqrt{15}-(x+y))}\color{black}{(x+y)=4-xy}$

Expanding the equation so that we have it written in the general quadratic equation format in terms of $(x+y)$, we get:

$(\pm\sqrt{15}-(x+y))(x+y)=4-xy$

$\pm\sqrt{15}(x+y)-(x+y)^2=4-xy$

$(x+y)^2\mp\sqrt{15}(x+y)+(4-xy)=0$

Now, this is a standard quadratic equation $[a(x+y)^2+b(x+y)+c=0]$ in terms of $(x+y)$ with:

$a=1,\,b=\mp\sqrt{15},\,c=4-xy$

Since we're told $x, y$ and $z$ are real numbers, the discriminant for the quadratic equation above must be greater than or equal to zero.

Therefore we have:

$(\mp\sqrt{15})^2-4(1)(4-xy)\ge 0$

Solving the inequality above for $xy$, we get:

$15-16+4xy\ge 0$

$4xy\ge 1$

$xy\ge \dfrac{1}{4}$

Therefore we can conclude that the minimum of $xy$ is $\dfrac{1}{4}$.