Suppose x,y, and z are real numbers.
Let:
A= 50x2+9y2+52+6xy+30xz+6yz
B= 25x2+41y2+5z2+40xy+20xz+26yz
C= 35x2+15y2+5z2+33xy+25xz+16yz
Given that C2=A*B, determine all possible values of x/y.
"AM GM": The Arithmetic Mean is >= The Geometric Mean
We know from AM GM that (A+B)/2 >= √(A*B)
So (A+B)/2 >= C
(A+B) >= 2C
A+B-2C >= 0
A+B = 75x^2+50y^2+10z^2+46xy+50xz+32yz
2C = 70x^2+30y^2+10z^2+66xy+50xz+32yz
A+B-2C = 5x^2+20y^2+ 0 -20xy+ 0 + 0
A+B-2C = 5x^2+20y^2-20xy
A+B-2C = 5(x-2y)^2 >= 0
Which unfortunately proves nothing.
But if C happened to be both the arithmetic mean and the geometric mean of A and B,
then we could say A+B=2C
and then we could say x-2y = 0 and x/y = 2.
So one solution for x/y is 2
For confirmation, I checked Wolfram Alpha but in addition to 2, it shows a second solution.
Edited on March 2, 2024, 11:56 am
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Posted by Larry
on 2024-03-02 11:56:27 |
Part of a solution
