Any inner product space $V$ is a normed space with norm $||f||=\sqrt{\langle f, f \rangle}$ 
(1) Show that any inner product space $V$ is a normed space with norm
  $||f||=\sqrt{\langle f, f \rangle}$, $\forall f\in V$.

Can someone please explain what kind of proof is required here? I've seen something like:

(2) If $X$ is a linear space with inner product $(.,.)$, then we can
  define a norm on $X$ by $||x|| = \sqrt{(x,x)}$. Thus, any inner
  product space is a normed linear space.

Where is the "proof" here? It seems like it is saying the same thing as $(1)$, so what is required of a proof to show the result of $(1)$?
 A: Since you have the norm axioms, verify that they each hold:


*

*$\|c \mathbf{x}\|=|c|\|\mathbf{x}\|$ for all scalars $c$ and vectors $\mathbf x$

*$\|\mathbf{x}+\mathbf{y}\|\le\|\mathbf{x}\|+\|\mathbf{y}\|$ for all vectors $\mathbf x, \mathbf y$

*$\|\mathbf x\|\ge 0$ with equality if and only if $\mathbf x=\mathbf 0$.
Just do this using $\|\mathbf x\|:=\sqrt{\langle\mathbf x,\mathbf x\rangle}$ with the scalars $c$ from whatever your underlying field is (probably $\mathbb R$ or $\mathbb C$).
A: The last condition to show that $|x| = \sqrt{\langle x,x \rangle}$ is  a norm (the other two are quite simple) should go like this: 
$$\begin{align}|x+y|^2 &= \langle x+y,x+y\rangle = |x|^2 + |y|^2 + 2\langle x,y\rangle \\ &\leq |x|^2 + |y|^2 + 2| x| \dot  \ |y| = (|x|+|y|)^2\end{align} $$
The question says that from an inner product, we may define the norm of a vector $x \in E$ by taking $|x|  = \sqrt{\langle x,x\rangle} $. Not every normed space $E$, however, comes from an inner product. This will only occur when the so called Parallelogram Law holds: 
$$|x+y|^2 + |x-y|^2 = 2(|x|+|y|)^2$$
For example the norm $|x|' = \sum |x_i| $ in $\mathbb{R}^n$ is not yielded from an inner product because the Parallelogram Law does not hold. 
If you have any question don't hesitate to ask on the comments. 
