# Show that $df_p(v)$ is maximum if and only if $v=\text{grad }f/|\text{grad }f|$

The gradient of a differentiable function $f:S\to \mathbb R$ is a differentiable map grad $f:S\to \mathbb R^3$ which assigns to each point $p\in S$ a vector grad $f(p)\in T_p(S)\subset \mathbb R^3$ such that $$\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v) \ \text{ for all } v\in T_p(S).$$

Show that

If you let $p\in S$ be fixed and $v$ vary in the unit circle $|v|=1$ in $T_p(S)$, then $df_p(v)$ is maximum if and only if $v=\text{grad }f/|\text{grad }f|$ (thus, $\text{grad }f(p)$ gives the direction of maximum variation of $f$ at $p$).

Here are some of my thoughts and doubts:
1. The $\Rightarrow$ direction, how can we make use if the fact that $df_p(v)$ is maximum? Do we take the derivative of $df$? If we need to make use of $\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v)$, does that mean we have to take the derivative of $\langle \operatorname{grad}{f(p)}, v\rangle_p$, but then I cannot see any obvious path to arrive at $v=\text{grad }f/|\text{grad }f|$ ?
2. The $\Leftarrow$ direction, the immediate implication from $v=\text{grad }f/|\text{grad }f|$ is $|v|=1$, but that is the initial assumption. Again I think we need to make use of $\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v)$ but I am not sure how to do so$?$

Those are the only tools that I could think of, I don't think we need to use directional derivative or any other operator, although I am not very sure.

Could somebody please give some clue on how to proceed$?$

Thanks for the help in advance

We have $\langle \text{grad} f(p),v\rangle = |\text{grad} f(p)|\cos\theta \leq |\text{grad} f(p)|$ if $|v| = 1$, and the upper bound is achieved for $v = \text{grad} f(p)/|\text{grad} f(p)|,$ making it the maximum.