Processing math: 100%

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 xy0, from the given equation xy+xz+yz=4, then we have:

(x+y)z4

So (x+y)216z2, and from the other given equation where x2+y2+z2=7:

7z2=x2+y2=(x+y)22xy(x+y)216z2.

Thus z2+16z27. AM-GM inequality formula tells us that cannot happen, because the minimum value of z2+16z22(z216z2)=2(4)=8, occurring when z2=4.

So we may assume that xy>0.

Let u=xy and v=x+y. Then we can write the equations as v22u2+z2=7, vz+u2=4.

Therefore v22(4vz)+z2=7, so (v+z)2=15.

But again from the AM-GM inequality we have: vz(12(v+z))2=154.

Therefore

vz+u2=4

u2=4vz

xy=u24154=14.

To see that the minimum value xy=14 is attained, check that if x=14(15+11), y=14(1511) and z=1215, then the original equations are satisfied, and xy=14.

To obtain those values for x, y and z, notice that the previous equations imply that to attain the minimum value 14 for u, we must have v=z=1215. In terms of x and y, this says that x+y=1215. Also, xy=14. This means that x and y are the roots of the quadratic equation λ21215λ+14=0, giving the values for x and y listed above.


No comments:

Post a Comment