[ $B_{r}=\left\{x \in\Bbb R^{n},|x| \leqq r\right\}$. The scalar product denoted by $(x, y)$ ]
Let $f: B_{r} \to\Bbb R^{n}$ be a continuous mapping satisfying on the boundary the condition $(f(x), x)>0 \quad \forall x,|x|=r$
I can't make an argument why this should implies $(f(x), x) \geqq K$ on $\partial B_{r}$ for a constant $K>0,$ and further the same inequality with a smaller constant in a neighbourhood of $B_{r}$, i.e. $$ (f(x), x) \geqq K / 2 \quad \forall x,|x| \in(\varrho, r], \quad \varrho<r $$
By continuity on a compact $\partial B_r$, your function attains its minimum there. Since it is strictly positive on $\partial B_r$, the minimum has to be positive. Call that minimum $K$ and you get the first assertion. The second one is because continuity on $B_r$ implies uniform continuity (again, because $B_r$ is a compact). Therefore, if the distance between points is less than some $\delta>0$ you can achieve the difference of the values of the function less than $K/2$. That implies that if you choose points sufficiently close to the boundary $\partial B_r$ ($\rho<\|x\|\le r$), the value of $(f(x),x)$ will be, say, within $K/2$ distance from its value at the boundary and, in particular, will be bounded below by $K/2$.