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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 x2+y2+z2=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, 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.


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