(k+2)2/(k2+ 1)= 1+((4k+3)/(k2+ 1))。 Suppose t = 4k+3/k 2+ 1.

Then: tk 2-4k+(t-3) = 0. Because k is a real number, the discriminant of this quadratic equation is greater than or equal to 0.

Therefore, △= 16-4t(t-3)>=0, (t-4) (t+1) < =0, so t

I hope my answer can help you ~