$$\nabla f(x,y,z)=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}$$
$\nabla$ is the gradient operator. $(\nabla f).\mathbf{r}$ is the rate of change of $f$ in the $\mathbf{r}$ direction.
I have to prove that a scalar multiple of $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}$ is the fastest rate of change of $f$ at any point.
Let $r$ be the unit vector $(a,b,c)$. $$a^2+b^2+c^2=1$$
$(\nabla f).\mathbf{r}=a.\frac{\partial f}{\partial x}+b\frac{\partial f}{\partial y}+c\frac{\partial f}{\partial z}$
How do we know this is maximized when $a=b=c$?
Best Answer
Recall that $$\nabla f(x)\cdot v=f'(x;v)$$ where this last term is the directional derivative of $f$ at $x$ with direction $v$. If you cannot see it directly, consider the function $$g(t)=f(x+tv)$$ and differentiate at $t=0$, using the usual definition of derivative, and then the chain rule. This gives you the "the fastest rate of change of $f$ at any point" idea. What you want to look at is $$F(v)=\langle \nabla f(x),v\rangle$$ under the constraint $\lVert v\rVert =1$. Informally, one usually says $$\langle \nabla f(x),v\rangle=\lVert \nabla f(x)\rVert \cos\theta$$
and since $-1\leq \cos\theta\leq 1$, this is made greatest when $\theta=0$, i.e. $v$ is $\nabla f(x)$, but normalized. More formally, we know by Cauchy Schwarz that $$\langle \nabla f(x),v\rangle\leq \lVert \nabla f(x)\rVert$$
and equality is attained iff $v$ is a (positive) multiple of $\nabla f(x)$. Thus, the constraint $\lVert v\rVert =1$ forces $$v=\frac{\nabla f(x)}{\lVert \nabla f(x)\rVert}$$