= Solution
The <gradient> of $x_3^2-x_1^2-x_2^2$ is $2(-x_1,-x_2,x_3)$. On the upper cone, its <Euclidean norm> is $2\sqrt2\,x_3$. Thus the <upward unit normal to an upper circular cone> is
$$
\boxed{n=\frac{(-x_1,-x_2,x_3)}{\sqrt2\,x_3},\qquad n_3=\frac1{\sqrt2}.}
$$
Put $x_h=(x_1,x_2)$ and $p_h=(p_1,p_2)$. The hypotheses imply $|x_h|=x_3$ and $|p_h|\le p_3$. The <Cauchy-Schwarz inequality> therefore gives
$$
p\cdot n=\frac{p_3x_3-p_h\cdot x_h}{\sqrt2\,x_3}
\ge\frac{p_3x_3-|p_h|\,|x_h|}{\sqrt2\,x_3}
=\frac{p_3-|p_h|}{\sqrt2}\ge0.
$$
This is also a direct instance of <self-duality of a second-order cone>: both $p$ and the chosen unit normal belong to the closed <Lorentz cone>.
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