<s>[INST] How find this minimum of the $(a-c)^2+(b-d)^2$ let $a,b,c,d\in R$,and such 
$$ab=c^2+4d^2=1$$
find the minimum of the value
$$(a-c)^2+(b-d)^2$$
My idea: since
$$(a-c)^2+(b-d)^2=a^2+b^2+c^2+d^2-2ac-2bd$$
I think  this inequality can use Cauchy-Schwarz inequality to solve it.Thank you
 [/INST] $f(a,b,c,d) = (a-c)^2 + (b-d)^2 \geq 0$, and $0$ is the minimum value of $f$. It occurs when:
$a = c$, and $b = d$. Thus:
$ab = cd = 1 = c^2 + 4d^2$. Solving this we get:
$a = c = \dfrac{5 \pm \sqrt{17}}{2}$, and
$b = d = \dfrac{2}{5 \pm \sqrt{17}}$
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