Thread: cauchy-schwarz problem

by considering suitable vectors in R4, show that only one choice of real numbers x,y,z satisfies :

3(x^2+y^2+z^2+4) - 2( yz+zx+xy) -4(x+y+z)=0

and find these numbers

Define:

Thus the equation turns into:

Note that:

Thus we have : or:

Consider:

Our equality is:

By Cauchy-Schwarz we have: with equality iff the vectors are colinear

So this must be the case and therefore is the only solution to this equation

Originally Posted by szpengchao

by considering suitable vectors in R4, show that only one choice of real numbers x,y,z satisfies :

3(x^2+y^2+z^2+4) - 2( yz+zx+xy) -4(x+y+z)=0

and find these numbers

If it were not for the vectors idea, we could have simply observed:



But PaulRS....awesome