there is 1>=a,b>=0 so a+b=1

prove that for two scalars u,v

$\displaystyle (au+bv)^{2}\leq au^{2}+bv^{2}$

i was told to prove it by proving that in bilenear u,v\in R

$\displaystyle q(u,v)=-(au+bv)^{2}+au^{2}+bv^{2}>=0$

i got there one minus so i am not sure if its positive

??