How prove there exsit $\xi\in (0,1)$ such $|f(\xi)|\le|f'(\xi)|$ 
Let $f:[0,1]\to \mathbb{R}$ be a differentiable function such that $f(1)=0$, Prove that there is $\xi\in(0,1)$, such that
$$|f(\xi)|\le|f'(\xi)|.$$

My idea:
I think we can prove there exsit $\xi\in (0,1)$ such
$$(f(\xi)-f'(\xi))(f(\xi)+f'(\xi))\le 0?$$ maybe we can consider function
$$F(x)=e^{\pm x}f(x)$$
But following can't works.
 A: It is trivial if $f$ has a zero in $(0,1)$, so suppose it doesn't.  Then $f$ can't change signs on $(0,1)$, and without loss of generality suppose $f$ is positive on $(0,1)$. Thus $\log(f)$ is defined on the interval, and $\lim\limits_{x\to 1-}\log(f(x)) = -\infty$, which implies by the mean value theorem that the derivative of $\log(f)$ takes on arbitrarily large negative values.  In particular, there exists $\xi\in(0,1)$ such that $\dfrac{f'(\xi)}{f(\xi)}\leq -1$.
A: Consider indeed $g:x\mapsto \mathrm e^xf(x)$.
We have $g(1)=0$ and $g'(x)=\mathrm e^x(f(x)+f'(x))$.
If for some $y\in(0,1)$, $f(y)>0$ then $g(y)>0$. We can consider that $f$ (and $g$) does not change sign on $(y,0)$. There exists a $\xi\in(y,1)$
such that $g'(\xi)<0$. Consequently $f(\xi)+f'(\xi)<0$. As $f(\xi)>0$, this becomes $-f'(\xi)>f(\xi)$ which gives the required result with strict inequality.
The reasoning is the same if we have $f(y)<0$ for $y\in(0,1)$. If there is no $y$ such that $f(y)>0$ or $f(y)$ then $f\equiv0$ and the result is trivial.
