 Self-duality of a second-order cone

For $(u,s),(v,t)$ in the second-order cone, the Cauchy-Schwarz inequality gives $u\cdot v+st\ge -\Vert u\Vert \Vert v\Vert +st\ge 0$. Thus the cone is contained in its dual cone. Conversely, if $s<\Vert u\Vert$ and $u\ne 0$, pairing $(u,s)$ with $(-u/\Vert u\Vert ,1)$ is negative. If $u=0$ and $s<0$, use $(0,1)$. Every point outside the cone therefore lies outside its dual, proving that it is a self-dual cone.