For $0<e<1<d$ and $n>2$, how can I prove the following inequality (if, in fact, it's true / possible)?
$$\frac{4n+4e}{4+n^2+4e+e^2}>\frac{4nd^2+4de}{4d^2+n^2d^2+2nde+e^2}$$
I tried setting $d=1, n=2$ to test the behaviour just beyond the limit, and it then becomes
$$\frac{4n+4e}{4+n^2+4e+e^2}=\frac{4n+4e}{4+n^2+4e+e^2}$$
Various tables I created suggest that the inequality is true... But I'm stumped on how to actually prove it.