(2) From the identityb α + c α = ( b α + 1 3 − c α + 1 3 ) ( b 2 α − 1 3 − c 2 α − 1 3 ) + b α + 1 c α + 1 3 ( b α − 2 3 + c α − 2 3 ) b^{\alpha}+c^{\alpha}=\left(\sqrt[3]{b^{\alpha+1}}-\sqrt[3]{c^{\alpha+1}}\right)\left(\sqrt[3]{b^{2 \alpha-1}}-\sqrt[3]{c^{2 \alpha-1}}\right)+\sqrt[3]{b^{\alpha+1} c^{\alpha+1}}\left(\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}\right) b α + c α = ( 3 b α + 1 − 3 c α + 1 ) ( 3 b 2 α − 1 − 3 c 2 α − 1 ) + 3 b α + 1 c α + 1 ( 3 b α − 2 + 3 c α − 2 )
we know that when α > 1 2 \alpha > \frac{1}{2} α > 2 1 b α + c α ⩾ b α + 1 c α + 1 3 ( b α − 2 3 + c α − 2 3 ) b^{\alpha}+c^{\alpha} \geqslant \sqrt[3]{b^{\alpha+1} c^{\alpha+1}}\left(\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}\right) b α + c α ⩾ 3 b α + 1 c α + 1 ( 3 b α − 2 + 3 c α − 2 )
Therefore,b c b α + c α + b c ⩽ b c b α + 1 c α + 1 3 ( b α − 2 3 + c α − 2 3 ) + b c ⩽ a α − 2 3 a α − 2 3 + b α − 2 3 + c α − 2 3 \begin{array}{l}
\frac{b c}{b^{\alpha}+c^{\alpha}+b c} \leqslant \frac{b c}{\sqrt[3]{b^{\alpha+1} c^{\alpha+1}}\left(\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}\right)+b c} \leqslant \\
\frac{\sqrt[3]{a^{\alpha-2}}}{\sqrt[3]{a^{\alpha-2}}+\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}}
\end{array} b α + c α + b c b c ⩽ 3 b α + 1 c α + 1 ( 3 b α − 2 + 3 c α − 2 ) + b c b c ⩽ 3 a α − 2 + 3 b α − 2 + 3 c α − 2 3 a α − 2
Similarly,a b a α + b α + a b ⩽ c α − 2 3 a α − 2 3 + b α − 2 3 + c α − 2 3 c a c α + a α + c a ⩽ b α − 2 3 a α − 2 3 + b α − 2 3 + c α − 2 3 \begin{array}{l}
\frac{a b}{a^{\alpha}+b^{\alpha}+a b} \leqslant \frac{\sqrt[3]{c^{\alpha-2}}}{\sqrt[3]{a^{\alpha-2}}+\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}} \\
\frac{c a}{c^{\alpha}+a^{\alpha}+c a} \leqslant \frac{\sqrt[3]{b^{\alpha-2}}}{\sqrt[3]{a^{\alpha-2}}+\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}}
\end{array} a α + b α + ab ab ⩽ 3 a α − 2 + 3 b α − 2 + 3 c α − 2 3 c α − 2 c α + a α + c a c a ⩽ 3 a α − 2 + 3 b α − 2 + 3 c α − 2 3 b α − 2
Adding the three inequalities, we get f ( α ) ⩽ 1 f(\alpha) \leqslant 1 f ( α ) ⩽ 1 α = − 1 \alpha=-1 α = − 1 α = 1 2 \alpha=\frac{1}{2} α = 2 1 f ( α ) = 1 f(\alpha)=1 f ( α ) = 1 − 1 < α < 1 2 -1<\alpha<\frac{1}{2} − 1 < α < 2 1 f ( α ) ⩾ 1 f(\alpha) \geqslant 1 f ( α ) ⩾ 1