Show that $ax+by=1 $ implies $\frac{1}{x^2+y^2} \leq a^2+b^2$ 
Show that for real positive x,y,a and b: $ax+by=1 \implies \frac{1}{x^2+y^2} \leq a^2+b^2$

I used to use contradiction but no results ;
We show that 
$$a^2+b^2 < \frac{1}{x^2+y^2} \implies ax+by \ne1$$
$$a^2+b^2 < \frac{1}{x^2+y^2} \iff (a^2+b^2)(x^2+y^2)<0 \iff (ax-by)^2-(ay-bx)^2 <0$$
I'm stuck here, I thought about proving it by putting something in a denominator which is not zero so it's not $1$.

Hint: 
For a nice geometric method, consider the circle $$x^2+y^2=\frac{1}{a^2+b^2}$$ and observe that any point on the line $$ax+by=1$$ is always outside of that circle, except for exactly one point. 
Recall that the perpendicular distance of a point $(x_0,y_0)$ from a line $Ax+By+C=0$ is $$d=\frac{\left | Ax_0 + By_0 + C \right |}{\sqrt{A^2+B^2}}$$
You may like to consider the distance of the line $ax+by=1$ from the origin, and you'll almost certainly observe that it has something to do with the circle I mentioned earlier.
