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:
$$z_1z_2 \geq \alpha^2k + \alpha(1-\alpha)(2k)+(1-\alpha)^2k \geq k (\alpha^2+2\alpha(1-\alpha)+(1-\alpha)^2)$$
$$z_1z_2 \geq k (\alpha + (1-\alpha))^2=k$$
I assume your definition of convexity is as follows: $K$ is said to be convex if whenever $x,y$ are in $K$, so too is $tx+(1-t)y$ for all $0\leq t\leq 1$.
If so, all you need to do is use induction.
Edit: Here's the answer:
Let $\sum_{i=1}^{n+1}t_{i}x_{i}$ be a convex combination. Then $\sum_{i=1}^{n+1}t_{i}x_{i}=\sum_{i=1}^{n}t_{i}x_{i}+t_{n+1}x_{n+1}$
where $t_{n+1}=1-\sum_{i=1}^{n}t_{i}$. We can write this as $$\left(\sum_{j=1}^{n}t_{j}\right)\left(\sum_{i=1}^{n}\frac{t_{i}}{\sum_{j=1}^{n}t_{j}}x_{i}\right)+\left(1-\sum_{i=1}^{n}t_{i}\right)x_{n+1},$$
which is a convex combination of two points in the set.
Best Answer
You just have to prove that for every $\alpha \in (0,1)$ and every pair $x_1$, $x_2$ $\in (1,5)$, $\alpha x_1 + (1-\alpha)x_2$ also lies in $(1,5)$.
Suppose without loss of generality that $x_1 \leq x_2$. Then for every $\alpha \in (0,1)$ (which implies that both $\alpha$ and $1-\alpha$ are positive)
$\alpha x_1 + (1-\alpha)x_2 \leq \alpha x_2 + (1-\alpha)x_2 = x_2$
and
$\alpha x_1 + (1-\alpha)x_2 \geq \alpha x_1 + (1-\alpha)x_1 = x_1$.
Hence we find that for every $\alpha \in (0,1)$, $\alpha x_1 + (1-\alpha)x_2 \in [x_1,x_2]$.
Hence for every pair $x_1$, $x_2$, any convex-combination of $x_1$, $x_2$ lies in $[x_1,x_2]$. And since $x_1$, $x_2$ $\in (1,5)$, every convex-combination of any two points in $(1,5)$ also lies in $(1,5)$.