How find this minimum of the $(a-c)^2+(b-d)^2$

Question

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

Answer 1

$f(a,b,c,d) = (a-c)^2 + (b-d)^2 \geq 0$, 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

$b = d = \dfrac{2}{5 \pm \sqrt{17}}$

Answer 2

There is another very simple solution without calculations.

Draw two figures in the plane: $xy = 1$ and $x^2 + 4 y^2 = 1$. These are a hyperbola and ellipse. The square of distance beetween them is the minimum of our value $(a-c)^2+(b-d)^2$. Check whether they have common points? $$\frac{1}{y^2} + 4 y^2 = 1 \Leftrightarrow 4y^4 -y^2 - 1 = 0$$ This square equation has a positive root. Hence, ellipse and hyperbola are intersect and the minimum value is equals to zero.