回復 5# satsuki931000 的帖子
第8題
\begin{align}
& \left( 2 \right) \\
& {{2}^{\frac{a}{2}+\frac{2}{b}}}+{{2}^{\frac{b}{2}+\frac{2}{c}}}+{{2}^{\frac{c}{2}+\frac{2}{a}}} \\
& \ge 3\sqrt[3]{{{2}^{\frac{a}{2}+\frac{2}{b}+\frac{b}{2}+\frac{2}{c}+}}^{\frac{c}{2}+\frac{2}{a}}} \\
& =3\sqrt[3]{{{2}^{\left( \frac{a}{2}+\frac{2}{a} \right)+\left( \frac{b}{2}+\frac{2}{b} \right)+}}^{\left( \frac{c}{2}+\frac{2}{c} \right)}} \\
& \ge 3\sqrt[3]{{{2}^{2+2+2}}} \\
& =12 \\
\end{align}
算幾第一個等號成立於a=b=c,第二個等號成立於a=b=c=2
\begin{align}
& \left( 3 \right) \\
& 4b-{{a}^{2}}={{2}^{\frac{a}{2}+\frac{2}{b}}} \\
& 4c-{{b}^{2}}={{2}^{\frac{b}{2}+\frac{2}{c}}} \\
& 4a-{{c}^{2}}={{2}^{\frac{c}{2}+\frac{2}{a}}} \\
& 4a+4b+4c-{{a}^{2}}-{{b}^{2}}-{{c}^{2}}={{2}^{\frac{a}{2}+\frac{2}{b}}}+{{2}^{\frac{b}{2}+\frac{2}{c}}}+{{2}^{\frac{c}{2}+\frac{2}{a}}} \\
\end{align}
左式最大值12,右式最小值12
故a=b=c=2