<s>[INST] 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 $1$.
 [/INST] Hint:
The magnitude of $(a+bi)(x+yi)$ is greater than the magnitude of the real part.
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