Thursday, June 4, 2015

Second Method of Solving IMO Optimization Contest Problem: Find minimum $xy$

Second Method of Solving 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.

The second method is the solution provided by a well-known mathematics retired professor from the University in the U.K.

He is a great and keen observer. He first noticed that the minimum value of $xy$ must be positive, because if $xy\leqslant 0$, from the given equation $xy+xz+yz=4$, then we have:

$(x+y)z\geqslant 4$

So $(x+y)^2\geqslant\dfrac{16}{z^2}$, and from the other given equation where $x^2+y^2+z^2=7$:

$7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \geqslant (x+y)^2 \geqslant \frac{16}{z^2}.$

Thus $z^2 + \dfrac{16}{z^2} \leqslant 7$. AM-GM inequality formula tells us that cannot happen, because the minimum value of $z^2 + \dfrac{16}{z^2}\ge 2\sqrt{\left(z^2\cdot \dfrac{16}{z^2} \right)}=2(4)=8$, occurring when $z^2 = 4$.

So we may assume that $xy>0$.

Let $u = \sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$.

Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$.

But again from the AM-GM inequality we have: $vz\leqslant\bigl(\frac12(v+z)\bigr)^2 = \frac{15}4.$

Therefore

$vz+u^2=4$

$u^2=4-vz$

$xy = u^2 \geqslant 4-\frac{15}4 = \frac14$.

To see that the minimum value $xy=\frac14$ is attained, check that if $x= \frac14(\sqrt{15} + \sqrt{11})$, $y= \frac14(\sqrt{15} - \sqrt{11})$ and $z= \frac12\sqrt{15}$, then the original equations are satisfied, and $xy = \frac14$.

To obtain those values for $x$, $y$ and $z$, notice that the previous equations imply that to attain the minimum value $\frac14$ for $u$, we must have $v=z=\frac12\sqrt{15}$. In terms of $x$ and $y$, this says that $x+y = \frac12\sqrt{15}$. Also, $xy = \frac14$. This means that $x$ and $y$ are the roots of the quadratic equation $\lambda^2 - \frac12\sqrt{15}\lambda + \frac14 = 0$, giving the values for $x$ and $y$ listed above.