Suppose that $K=\{(x,y,z)\in \mathbb R^3|x\geq 0,y\geq 0, xy\geq z^2\}$.
Let $K_0=\{x\in\mathbb R^3|\langle x,k\rangle\leq 0\forall k\in K\}$.
What would be a description of the set $K_0$? I don't have the slightest idea how to attack such a question. Any help will be appreciated. Here, $\langle\cdot,\cdot\rangle$ denotes the usual dot-product in $\mathbb R^3$, of course.
The problem tag is convex-analysis, so we analyze it. Let $x=(x_1,x_2,x_3)$ be a vector in $\Bbb R^3$ such that
(#) $x_1k_1+x_2k_2+x_3k_3=\langle x,k\rangle\le 0$ for each vector $k=(k_1,k_2,k_3)\in K$.
Since $k_3$ may take all values from the segment $[-\sqrt{k_1k_2},\sqrt{k_1k_2}]$, the condition (#) holds iff
(##) $x_1k_1+x_2k_2\le -|x_3|\sqrt{k_1k_2}\le 0$ for each vector $k=(k_1,k_2,k_3)\in K$.
Since vectors $(1,0,0)$ and $(0,1,0)$ belong to the set $K$, we obtain that both $x_1$ and $x_2$ are non-positive. Moreover, if $x_i=0$ for $i=1$ or $i=2$ then if we fix $k_{3-i}=1$ and tend $k_i$ to infinity, we obtain that $x_3=0$. By AM-GM, $|x_1|k_1+|x_2|k_2\ge 2\sqrt{|x_1||x_2|k_1k_2}$, so to satisfy (##) if suffices to assure $|x_3|\le 2\sqrt{|x_1||x_2|}$. From another side, if both $x_1$ and $x_2$ are non-zero, by putting $k_1=|x_2|$ and $|k_2|=|x_1|$ we see that $|x_3|\le 2\sqrt{|x_1||x_2|}$.
Final answer: $K_0=\{(x,y,z)\in \mathbb R^3|x\leq 0,y\leq 0, 2xy\geq z^2\}$.