How to show that $|\nabla f|\ge \sqrt 2$?

Question

Let $[0,1]^2\subset U\subset \mathbb R^2$ where $U$ is open, and let $f: U\to \mathbb R$ be a differentiable function with $f(0,0)=3$ and $f(1,1)=1$. Prove that $|\nabla f|\ge \sqrt 2$ somewhere in $U$.

The only idea I had is to apply Taylor's formula but it only gives something like $-2=f(1,1)-f(0,0)=f_x(1,1)+f_y(1,1)$, which holds at a point, and I guess "somewhere" means "on a subset". Moreover, I don't know how to get hold of $f_x^2+f_y^2=||\nabla f|| ^2$.

Any hints please? (If I only use hints as opposed to reading a solution I hope I will learn more.)

Answer 1

Hint: Mean value theorem in one variable, applied to $f(x,y)$ on the line segment connecting $(0,0)$ and $(1,1)$.

You also might have a theorem in your book which gives a version of the mean value theorem that applies in the setting of your question. It will basically derive from the above idea.

Answer 2

By MVT we have that $\exists c=a+(b-a)t\quad t\in(0,1)$

$$f(b)-f(a)=\frac{\partial f(c)}{\partial \vec v}|b-a|$$

with $\vec v=\frac{b-a}{|b-a|}$ and for $f$ differentiable we have

$$\frac{\partial f(c)}{\partial \vec v}=\langle \nabla f(c),\vec v\rangle=\frac1{|b-a|}\langle \nabla f(c), b-a\rangle$$

therefore

$$f(b)-f(a)=\langle \nabla f(c), b-a\rangle$$

that is

$$2=f(0,0)-f(1,1)=\langle \nabla f(c), (-1,-1)\rangle=|\nabla f(c)|\sqrt 2 \cos \theta$$

and finally

$$|\nabla f(c)|=\frac{\sqrt 2}{\cos \theta}\ge \sqrt 2$$