<s>[INST] 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?
 [/INST] The minimal value of the parabola
$$
a^2+b^2=\left( \frac 1 2-2\,b \right) ^{2}+{b}^{2}=\frac 1 4-2\,b+5\,{b}^{2}
$$
is at its vertex $b=\dfrac{1}{5}$ and it equals exactly $\dfrac{1}{20}.$
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