given four real numbers $a, b, c, d$
Ji Chen gave a nice result on.AoPS $$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq\frac{7}{9}ab- \frac{7}{20}\left ( c^{2}+ 4d^{2} \right )$$
The original problem consists a condition $ab= c^{2}+ 4d^{2}= 4$ and hides the rightside of the earlier inequality, I wondered that how he could find this.. of course I searched in AoPS because of him but I also found another interesting work, which made me sense.
He (AoPS/@ye109) claimed that $$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq\frac{2\left ( k^{4}- 4 \right )}{3k^{2}}ab+ \frac{k^{4}- 4}{k^{4}- 16}\left ( c^{2}+ 4d^{2} \right )$$ with the best $k\cong 1.672955253$ is a real root of $k^{12}- 28k^{8}- 36k^{6}- 128k^{4}+ 1024$
That guy is very good at making second solution, we just need explain his solution from Ji Chen's 1st one, if Ji Chen used discriminant to find $7/9, -7/20$ then how could he use discriminant to find $\frac{2\left ( k^{4}- 4 \right )}{3k^{2}}, \frac{k^{4}- 4}{k^{4}- 16},$ maybe there could exist discriminant for polynomials not only for constants. What's your idea ? Who can give me a way of thinking ? Thanks a real lot !
$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq Uab-V \left ( c^{2}+ 4d^{2} \right )$$ (for all reals $a,b,c,d$)
precisely when $$ V \geq 0$$ and $$ |U| \leq \frac{4V}{\sqrt{(1+V)(1+4V)}} $$
In turn, the largest $U$ is rational when $$ V = \frac{m^2 - n^2}{4n^2 - m^2} \geq 0$$ where we get $$ |U| \leq 4\frac{|m^2-n^2|}{3mn} $$
In particular, when we take $n=3,m=4$ we get $V= 7/20$ and the largest $|U|$ that still gives positive semidefinite is $7/9$
The positive semidefinite matrix with extremal $|U|$
using $2n > m > n,$ $$ \left( \begin{array}{cccc} 1&\frac{2m^2-2n^2}{3mn}&-1&0\\ \frac{2m^2-2n^2}{3mn}&1&0&-1\\ -1&0&\frac{3n^2}{4n^2-m^2}&0\\ 0&-1&0&\frac{3m^2}{4n^2-m^2}\\ \end{array} \right) $$