DreamTeam wrote: Let $ x\in \mathbb R$. Find the minimum and maximum values of the expresion: $ E = \dfrac{(1 + x)^8 + 16x^4}{(1 + x^2)^4}$ $ E=\dfrac{(1 + x)^8 + 16x^4}{(1 + x^2)^4}=(1+\frac{2x}{1+x^{2}})^{4}+(\frac{2x}{1+x^{2}})^{4}$ $ a=\frac{2x}{1+x^{2}}\leq 1$---> $ E=(1+a)^{4}+a^{4}\leq 2^{4}+1^{4}=33$ --->$ max(E)=33$ ($ x=1$) $ =(1+a)^{4}+a^{4}=(0.5+(a+0.5))^{4}+((a+0.5)-0.5)^{4}=0.5^{4}+4*0.5^{3}(a+0.5)+6*0.5^{2}(a+0.5)^{2}+4*0.5(a+0.5)^{3}+(a+0.5)^{4}+0.5^{4}-4*0.5^{3}(a+0.5)+6*0.5^{2}(a+0.5)^{2}-4*0.5(a+0.5)^{3}+(a+0.5)^{4}=2*0.5^{4}+2*0.5^{2}(a+0.5)^{2}+2(a+0.5)^{4}=2(a+0.5)^{4}+\frac{1}{2}(a+0.5)^{2}+\frac{1}{8}\geq\frac{1}{8}$- > $ min(E)=\frac{1}{8}$ ($ a=-0.5$ and $ x=-2+\sqrt{3}$)

Let $ x = \tan \theta$, we have $ E = (\cos \theta + \sin \theta)^8 + 16\sin ^ 4 \theta \cos ^ 4 \theta$, and let $ t = \sin 2\theta$, we obtain $ E = (1 + t)^4 + t^4$ with $ - 1\leq t\leq 1$.

Just to point out that $2^4 + 1^4 = 17$, not $33$.