Show that if $2x + 4y = 1$ where x and y are real numbers. Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$
I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!
If $x=2$ and $y=4$, then $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$ iff $$\frac{x}{2}=\frac{y}{4}$$, then $$20\ge(2x+4y)^2$$
$$-4.47\le2x+4y\le4.47$$

The distance from the circle's center ($ x^2+y^2 = \frac{1}{20}$) to the line $2x + 4y = 1$ is given by
$$
d = \frac{1}{\sqrt{2^2+4^2}} = \sqrt{\frac{1}{20}}
$$
