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?
 A: 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:
$$a^2+b^2=\frac{1}{5}(1^2+2^2)(a^2+b^2)\geq\frac{(a+2b)^2}{5}=\frac{1}{20}.$$
A: An alternative approach is to use AM-QM inequality as
$$\left(\frac{2a + b + b + b +b }{5}\right)^2 \le \frac{4a^2+b^2+b^2+b^2+b^2}{5} \implies a^2+b^2 \ge \frac{1}{20}.$$
A: 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
$$x^2+y^2 \ge \frac{1}{20}$$
A: Since
$$
a=\frac12-2b
$$
we have
$$
\begin{align}
a^2+b^2
&=\frac14-2b+5b^2\\
&=5\left(b-\frac15\right)^2+\frac1{20}
\end{align}
$$
A: 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 $\dfrac{1}{20}.$
A: this is equivalent to $$(10a-1)^2\geq 0$$
by inserting $$b=\frac{1}{4}-\frac{1}{2}a$$ in the left Hand side and factorizing!
and the term above is $$\frac{5}{4}a^2+\frac{1}{16}-\frac{1}{4}a-\frac{1}{20}$$
