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?

An alternative approach is to use AM-QM inequality as
$$\left(\frac{2a + b + b + b +b }{5}\right)^2 \le \frac{4a^2+b^2+b^2+b^2+b^2}{5} \implies a^2+b^2 \ge \frac{1}{20}.$$
