I disagree with your second attempt, I don't understand it completely though, it's a bit ambiguous what you're saying, but the way I interpret it sounds wrong, but your first attempt is definitely wrong, you have to show that inequality holds for the coordinates of all vectors that lie in the line segment between each two vectors in the set.
Suppose that for some arbitrarily chosen $0\leq \alpha \leq 1$ we write:
$(z_1,z_2)=\alpha (x_1,x_2) + (1-\alpha) (y_1,y_2)$
Does $z_1 \cdot z_2 \geq k$ hold?
Remember that $x_1x_2 \geq k$ and $y_1y_2\geq k$.
Addendum:
This is how I can prove what you want, I don't use $\displaystyle \frac{a}{b} + \frac{b}{a} \geq 2$ in my proof, but since I'm using AM-GM inequality, I guess that with a suitable change of variables you might find a way to write what you want by using that particular inequality (which is implied by AM-GM):
As you said, we have:
$z_1 = \alpha x_1 + (1-\alpha)y_1$
$z_2 = \alpha x_2 + (1-\alpha)y_2$
Just by doing some simple algebraic manipulations, we get:
$z_1 z_2 = \alpha^2 x_1x_2 + \alpha(1-\alpha)x_1y_2 + \alpha(1-\alpha)x_2y_1 + (1-\alpha)^2y_1y_2$
Now use your hypotheses to get:
$z_1z_2 \geq \alpha^2 k + \alpha(1-\alpha)(x_1y_2+x_2y_1) +(1-\alpha)^2k$
But by using AM-GM inequality we have:
$x_1y_2+x_2y_1 \geq 2 \sqrt{x_1x_2}\sqrt{y_1y_2}\geq 2k$
Therefore, we get:
$$z_1z_2 \geq \alpha^2k + \alpha(1-\alpha)(2k)+(1-\alpha)^2k \geq k (\alpha^2+2\alpha(1-\alpha)+(1-\alpha)^2)$$
$$z_1z_2 \geq k (\alpha + (1-\alpha))^2=k$$
This set is the epigraph of the $2$-norm, which is a convex function. Hence, this set is convex.
By the way, this set is called the "second-order cone" or the "ice-cream cone". (More precisely, this is the ice-cream cone in $\mathbb R^3$.)
Best Answer
Say $f(x)=\frac{1}{x}$, so our set is $\{(x,y): y\geq f(x)\}$. We know that $f(x)$ is convex, that is for $0\leq\theta\leq 1$ $$f(\theta x+(1-\theta)y)\leq \theta f(x)+(1-\theta)f(y).$$
Now take $(x_1,y_1)$ and $(x_2,y_2)$ in the set. In other words $y_1\geq f(x_1)$ and $y_2\geq f(x_2)$. For $0\leq\theta\leq 1$ $$\theta y_1+(1-\theta)y_2\geq\theta f(x_1)+(1-\theta)f(x_2)\geq f(\theta x_1+(1-\theta)x_2)$$ so it is in the set.