Đặt $x=\tan t$. Suy ra
$\frac{1+x^4}{1+x^2}=\frac{1+\tan^4 t}{1+\tan^2 t}=\frac{1+\tan^4 t}{\frac{1}{\cos^2 t}}=\cos^2t(1+\tan^4 t)=\cos^2t+\frac{\sin^4t}{\cos^2t}$
$=\cos^2 t+\frac{(1-\cos^2 t)^2}{\cos^2t}=\cos^2 t+\frac{1-2\cos^2t+\cos^4t}{\cos^2t}=2\cos^2t +\frac{1}{\cos^2t}-2$
Aps dụng BĐT Cô-si:
$\frac{1+x^4}{1+x^2} \ge 2\sqrt{2\cos^2t .\frac{1}{\cos^2t}}-2 =2\sqrt 2 -2 > \frac12$.
BĐT $\frac{1+x^4}{1+x^2} \le 1$ sai vì cho $x=2$ thì
$\frac{1+x^4}{1+x^2}=\frac{17}{5}>1$.