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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 $2\sqrt{2}{x}_3$. Thus the upward unit normal to an upper circular cone isPut $x_h=(x_1,x_2)$ 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 givesThis 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.

A curved feasible set can have the same certificate structure. ConsiderWrite $x=(t,z{)}^T$ and use the second-order cone $K=\{(s_0,s_1,s_2):s_0\ge \sqrt{s_1^2+s_2^2}\}$. In this coordinate ordering,Self-duality of a second-order cone makes $K^{\ast }=K$. The equality $A^{T}y=c$ forces $y_0=1$ and $y_2=0$, leaving $-1\le y_1\le 1$. Its dual objective function is $b^{T}y=-y_1$, maximized at $y=(1,-1,0{)}^T$. Thus every feasible point satisfies $t\ge 1$, and $(t,z)=(1,0)$ attains that bound: $\boxed{p_{\ast }=d_{\ast }=1}$.

The zero dot product is $(1,1,0)\cdot (1,-1,0)=0$. Individual coordinate products need not be nonnegative here: the cancellation $1-1=0$ occurs across coordinates. The valid complementary slackness condition is cone-level orthogonality, rather than the coordinatewise interpretation of the nonnegative orthant.

On $x_3^2=x_1^2+x_2^2$ with $x_3>0$, the gradient of the defining function gives a normal proportional to $(-x_1,-x_2,x_3)$. Its Euclidean norm is $\sqrt{2}{x}_3$, proving the displayed normalization. This upward normal lies on the boundary of the Lorentz cone, so self-duality of a second-order cone makes its pairing with every point of the closed upper cone nonnegative.