alternate proof of fixed point theorem There is a famous application of fixed point theorem as follows :

If $f:\mathbb{R}^n\to\mathbb{R}^n$ be a continuous function such that
$\langle f(x),x\rangle\geq0$ for all $|x|=R>0$ , then $f(x_0)=0$ for
some $|x_0|\leq R$ .

My question is , if we assume the validity of the above result , then from there can we prove the fixed point theorem for the closed ball $B(0,R)$ ?
If not , we have $f(x)\neq x$ for all $|x|\leq R$ . By contrapositive of the result one has $\langle f(x_0)-x_0,x_0\rangle <0$ for some $|x_0|=R$ , i.e. $\langle f(x_0),x_0\rangle<R^2$ . But I can't arrive at a contradiction from there .
Any help is appreciated , regards .
 A: Let $f : B(0,R) \to B(0,R)$ be continuous. Define
$$g : B(0,R) \to \mathbb R^n, g(x) = x - f(x) ,$$
$$r : \mathbb R^n \to B(0,R), r(x) = \begin{cases} x & \lvert x \rvert \le R \\ R\frac{x}{\lvert x \rvert} & \lvert x \rvert \ge R \end{cases}$$
Then
$$F = g \circ r : \mathbb R^n \to \mathbb R^n$$
is a continuous map. We want to show that $\langle F(x), x \rangle \ge 0$ for $\lvert x \rvert > R$; this implies that $F(x_0) = 0$ for some $x_0 \in B(0,R)$. But for such $x_0$ we have $F(x_0) = g(x_0) = x_0 - f(x_0)$, thus $f(x_0) = x_0$.
Setting $y = R\frac{x}{\lvert x \rvert}$, we get
$$\langle F(x), x \rangle = \frac{\lvert x \rvert}{R}\langle g(y), y \rangle . $$
Thus it suffices to show that $\langle g(y), y \rangle \ge 0$ if $\lvert y \rvert = R$. Since $f(y) \in B(R,0)$, we get
$$\langle f(y), y \rangle \le \lvert f(y) \rvert \cdot \lvert y \rvert = \lvert f(y) \rvert \cdot R \le R^2$$
and therefore
$$\langle g(y), y \rangle = \langle y - f(y), y \rangle =\lvert y \rvert ^2 - \langle f(y), y \rangle = R^2 - \langle f(y), y \rangle \ge R^2 - R^2 = 0 .$$
