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
??