<s>[INST] Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$ Let $a, b, c, d$ be real numbers. I need to prove this innocent inequality
$$
(a-d)^2+(b-c)^2\geq 1.6
$$
if $a^2+4b^2=4$ and $cd=4$.
I was told that there exist nice and sweet elementary solutions.
 [/INST] it is supplementary to Alex's method which more like high school method:
The idea is same,we use two parallel lines to calculate. let slop is $k=-\tan{ \alpha}, 0<\alpha <\dfrac{\pi}{2}$,
the tangent line to ellipse $x^2+4y^2=4$ is :$y=kx+b_1 \to x^2+4(kx+b_1)^2=4, \Delta=0,\implies b_1^2=4k^2+1$
the tangent line to  Hyperbola $xy=4$ is:$y=kx+b \to x(kx+b)=4, \Delta=0,\implies k=-\dfrac{b^2}{4^2}$
$b_1=\sqrt{\dfrac{b^4}{64}+1},\cos{\alpha}=\dfrac{1}{\sqrt{1+\tan^2{\alpha}}}=\dfrac{4^2}{\sqrt{b^4+4^4}}$
the distance between two lines is: $D=(b-b_1)\cos{\alpha}=\left(b-\sqrt{\dfrac{b^4}{64}+1}\right)\dfrac{4^2}{\sqrt{b^4+4^4}}=\dfrac{2(8b-\sqrt{b^4+64})}{\sqrt{b^4+4^4}}=\dfrac{2(64b^2-b^4-64)}{\sqrt{b^4+4^4}(b+\sqrt{b^4+64})}=\dfrac{2(960-(b^2-32^2)^2)}{\sqrt{b^4+4^4}(b+\sqrt{b^4+64})}$
the numerator is Parabola, the peak is at $b^2=32,b=4\sqrt{2}$
the denominator is mono increasing function, and it is lead by $b^4$, so it will cause the peak of $D$ move left strongly. The $D(b)$ has two zero points at about $b=1,8$, so the peak point should be around $3$ or $4$( it is guess).
$D(3)=1.3,D(4)=1.25 \implies D^2=1.69>1.6$  so we have at least one solution to show the square of distance of two points $(a,b),(c,d) >1.6 $
if we try $D(3.5)=1.76$, so if we set min is $1.7$,it is much hard to high school student.
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