Proving $\{(x_1, x_2) \in \mathbb{R}^2_{+} \mid x_1 x_2 \geq 1\}$ is convex

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.

Let $(x_1,y_1)$ and $(x_2, y_2)$ in our set, and $\alpha \in [0,1]$. We need to show that $\alpha (x_1,y_1) + (1-\alpha) (x_2, y_2)$ is in our set, i.e. that $$\alpha^2 x_1y_1 + (1-\alpha)^2 x_2y_2 + \alpha(1-\alpha)(x_1y_2+x_2y_1) \geq 1$$

Hint: notice that $x_1y_1 \geq 1, x_2y_2 \geq 1$ and prove that $x_1y_2 + x_2y_1 \geq 2$ to conclude.