Given $\{x \in \mathbb{R}^2_+: x_1 x_2 \geq 1\}$, find expression of supporting hyperplane $\{x \in \mathbb{R}^2_+: x_1 x_2 \geq 1\}$ is difficult to sketch because the boundary of this set is the set $x = (t, 1/t)$, $t > 0$ and I don't have a good idea where this supporting hyperplane is located
But nonetheless recall that a supporting hyperplane satisfies the condition $\{x | a^Tx = a^Tx_o\}$
Here we know that a boundary point is $(t, 1/t)$, how do we use this information to find the supporting hyperplane of the form $x_2 - x_2^o = m(x_1 - x_1^o)$?
Note that the correct solution seems to be $x_1 / t^2 + x_2 = 2/t$

It's somewhat simple in this case since the boundary of this set is the graph of the function $f(x)=1/x$ $(x>0)$; i.e., $x_2=1/x_1$. The slope of the tangent is therefore just given by the derivative of $f$; that is, $f'(x)=-1/x^2$.
Thus at any $t>0$, the supporting hyperplane is
$$x_2 = f(t) + ( x_1 - t ) f'(t) = 1/t - ( x_1 - t ) / t^2 = 2/t - x_1 / t^2.$$
Move the $x_1$ term over and you're done.
