<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] We need to prove that $$a^2+b^2\geq\frac{(2a+4b)^2}{20}$$ or
$$(2a-b)^2\geq0.$$
Also, we can use C-S:
$$a^2+b^2=\frac{1}{5}(1^2+2^2)(a^2+b^2)\geq\frac{(a+2b)^2}{5}=\frac{1}{20}.$$
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