Inner Product Space | find $\langle u+2v, u-2v\rangle$ 
Let $u$ and $v$ be vectors in inner product space, and suppose that $\|u\| \ge 3 $ and $\|v\| = 5$ find $\langle u+2v, u-2v\rangle$

I have no idea how to solve this. The only thing i know is; $\|u\| = 
\sqrt{\langle u,u\rangle } $
 A: Assuming real vectors: $\bar{u} = u$ and $\bar{v} = v$
$\langle u + 2\,v\,,\,u-2\,v\rangle = \langle u \,,\,u-2\,v\rangle + 2\,\langle v\,,\,u-2\,v\rangle = \langle u-2\,v \,,\,u\rangle + 2\,\langle u-2\,v\,,\,v\rangle = \\\langle u\,,\,u\rangle - 2\,\langle v\,,\,u\rangle + 2\,\langle u\,,\,v\rangle - 4\,\langle v\,,\,v\rangle = ||u||^2-4\,||v||^2 \geq -91$
Inner Product
A: As a rough intuitive idea think of the special case of an inner product space that is $\mathbb{R}$ with usual multiplication, i.e. $\langle x , y \rangle = x \cdot y$. Note that the product you are supposed to evaluate has a very particular form, i.e. it is of the form
$$
\langle x + y, x - y \rangle,
$$
which in our real-number setting will  be $(x-y)\cdot(x+y)$ and this can be calculated by the binomial formulas as $x^2 - y^2$, since the cross terms will cancel. If you perform the calculations, you will see that the same holds true in the case of your general inner product space, i.e. in this particular case, you only need to know what $\langle x, x \rangle$ (corresponds to $x^2$ above) and $\langle y, y \rangle$ (corresponds to $y^2$ above) are, and these are related to the norms.
To make a long story short: simplify the expression via bilinearity and expect to find that the "cross-terms" will cancel, and then plug in what you know about the norms of $u$ and $v$.
Hope that helps!
Andre
