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:
$$z_1z_2 \geq \alpha^2k + \alpha(1-\alpha)(2k)+(1-\alpha)^2k \geq k (\alpha^2+2\alpha(1-\alpha)+(1-\alpha)^2)$$
$$z_1z_2 \geq k (\alpha + (1-\alpha))^2=k$$
You need to show that $f(x) \leq \alpha$, where $x$ is chosen as convex combination of the points $x_1$ and $x_2$, i.e. $x=(1-\lambda)x_1 + \lambda x_2$.
Now,
\begin{align*}
f(x)& =f((1-\lambda)x_1 + \lambda x_2) \\
&\leq (1-\lambda)f(x_1) + \lambda f(x_2) \;\; \text{(using convexity of $f(\cdot)$)}\\
& \leq(1-\lambda)\alpha + \lambda \alpha \;\;\text{(using epigraph definition)}\\
& = \alpha
\end{align*}
Thus, epi $f$ is convex. q.e.d
Best Answer
You have to show that if $x_1,x_2$ satisfy the inequality, then so does $\lambda x_1+(1-\lambda)x_2$, for any $\lambda \in (0,1)$. So the goal is to show that $$\|A(\lambda x_1+(1-\lambda)x_2)+b\|\leq c^T(\lambda x_1+(1-\lambda)x_2)+d,$$assuming that $\|Ax_i+b\|\leq c^Tx_i+d$, for $i=1,2$, which is not quite what you have there. We do not think of "convexity on each side". One proceeds as follows: $$\begin{align} \|A(\lambda x_1+(1-\lambda)x_2)+b\| & = \|\lambda (Ax_1+b) + (1-\lambda )(Ax_2+b)\| \\ &\leq \lambda \|Ax_1+b\| + (1-\lambda)\|Ax_2+b\| \\ &\leq \lambda(c^Tx_1+d) + (1-\lambda)(c^Tx_2+d) \\ &= c^T(\lambda x_1+(1-\lambda)x_2) + d.\end{align}$$