I tried taking a convex combination of $x$ and $y$ (given by $(a,b)$ and $(s,t)$ respectively) but the resulting expression had two terms which I could not get rid of or substitute with anything useful:
$$k(1-k)bs$$ $$k(1-k)at$$
I suspect I just have to manipulate the algebraic expression so that I can use the conditions $ab >=1$ and $st >=1$ but I haven't been able to do this.
Let, $(x_1,y_1),(x_2,y_2)\in A$
Again, let $(x_3,y_3)=\lambda (x_1,y_1)+(1-\lambda)(x_2,y_2)$ where $0\le \lambda\le 1$
Then, $x_3y_3=\lambda^2x_1y_1+(1-\lambda)^2x_2y_2+\lambda(1-\lambda)(x_2y_1+x_1y_2)\ge \lambda^2+(1-\lambda)^2+2\lambda(1-\lambda)=1$
and we are done!
An alternative is to use Proof that a function is convex if and only if its epigraph is convex.
Here since $y>0$ (in particular $x\neq 0$) the function $y=f(x)=\frac 1x$ is well defined and it is convex since $f''(x)=\frac 2{x^3}>0$ for $x>0$.
