Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$ Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$
The only idea I have is that I could apply Cauchy-Schwartz but i don't see how, any hints?

Using coordinate geometry
The perpendicular distance of $2x+4y=1$ from the origin is given by
$$d=\frac{1}{\sqrt{2^2+4^2}}=\frac{1}{\sqrt{20}}$$
By $\sqrt{x^2+y^2} \ge d$ gives
$$x^2+y^2 \ge \frac{1}{20}$$
