If a,b,c,x,y,z are positive and x2+y2+z2=36, and ax+by+cz=30, compute
a+b+cx+y+z
well.... if this is true for all a,b,c,x,y,z, then I can choose (x,y,z)=2√3(1,1,1)(a,b,c)⋅(x,y,z)=302√3(a+b+c)=30(a+b+c)=5√3x+y+z=6√3a+b+cx+y+z=5√36√3=56