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⩽0, from the given equation xy+xz+yz=4, then we have:
(x+y)z⩾4
So (x+y)2⩾16z2, and from the other given equation where x2+y2+z2=7:
7−z2=x2+y2=(x+y)2−2xy⩾(x+y)2⩾16z2.
Thus z2+16z2⩽7. AM-GM inequality formula tells us that cannot happen, because the minimum value of z2+16z2≥2√(z2⋅16z2)=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 v2−2u2+z2=7, vz+u2=4.
Therefore v2−2(4−vz)+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=4−vz
xy=u2⩾4−154=14.
To see that the minimum value xy=14 is attained, check that if x=14(√15+√11), y=14(√15−√11) and z=12√15, 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=12√15. In terms of x and y, this says that x+y=12√15. Also, xy=14. This means that x and y are the roots of the quadratic equation λ2−12√15λ+14=0, giving the values for x and y listed above.
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