maximum value of $ac+bd$ 
If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is 

what i try
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$
$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$
but answer is $3$
How do i get right answer help me please

You are correct that the maximum value is $2\sqrt 2<3$. As another answer (now deleted; it applied AM-GM separately to $a^2+c^2$ and $b^2+d^2$) points out, you can easily get an upper bound of $3$ using AM-GM, but this bound can't actually be attained, so isn't the correct maximum. 
For equality in the AM-GM bound, you would need $a=c$ and $b=d$, but this is clearly impossible since $a^2+b^2\neq c^2+d^2$.
