<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] Using $a^2 + 4b^2 = 4$ and $cd = 4$, we have
\begin{align*}
 &(a - d)^2 + (b - c)^2 + \frac12(a^2 + 4b^2 - 4) - \frac{2\sqrt 2}{3}(cd - 4)\\[6pt]
 =\,& \frac16(3a - 2d)^2 + \frac13(3b - c)^2 + \frac13(d - c\sqrt 2)^2 + \frac{8\sqrt 2}{3} - 2 \\[6pt]
 \ge\, &\frac{8\sqrt 2}{3} - 2\\
 >\,& \frac85.
\end{align*}
We are done.
</s>