Show that the maximum of $v^TAv$ over all $v\in \mathbb{R}^p$ with $v^TBv=1$ is the same as $\text{max}_{v\in\mathbb{R}^p,v\neq 0}\frac{v^tAv}{v^TBv}$

Question

I'm doing an exercise which I don't understand:

Let $A,B$ be $p\times p$ symmetric, and let $B$ be positive definite. Show that the maximum of $v^TAv$ over all $v\in \mathbb{R}^p$ with $v^TBv=1$ is the same as $$\text{max}_{v\in\mathbb{R}^p,v\neq 0}\frac{v^tAv}{v^TBv}$$

So, I'm not sure I completely understand what the problem asks me to do. What I think it asks me to do is of the form:

Show that the one + one is equal to 2.

And I would just write $1+1=2$.

I'm quite confused what they're asking me to do.

If anyone could help me, that would be great.