## 2009年10月1日 星期四

### AM-GM 不等式的幾個證明（四）

$G_{k+1}=\sqrt[k+1]{x_1x_2\cdots x_{k+1}}$。由歸納假設可知

$\frac{{x_1x_{k+1}}}{{(G_{k+1})^2}} + \frac{{x_2 }}{{G_{k+1}}}+\cdots+\frac{{x_{k+1}}}{{G_{k+1}}}\ge k\cdot\sqrt[k]{{\frac{{x_1x_2\cdots x_{k + 1} }}{{(G_{k+1})^{k+1}}}}} = k$

$\frac{{x_1 + x_{k + 1} }}{{G_{k + 1} }} - \frac{y}{{(G_{k + 1} )^2 }} \ge 1 \Leftrightarrow (G_{k + 1} - x_1 )(G_{k + 1} - x_{k + 1} ) \ge 0$

$R=\frac{\sqrt[k]{x_1x_2\cdots x_k}}{x_{k+1}}$，則 $0，故此

\100dpi \begin{align*} x_1+x_2+\cdots+x_{k+1}&\ge(kR+1)x_{k+1}\\ &=\left[\left(1-R^{\frac1{k+1}}\right)\left(1+R^{\frac1{k+1}}+\cdots+R^{\frac{k}{k+1}}\right)\right]x_{k+1}\\ &>\left[\left(1-R^{\frac1{k+1}}\right)(k+1)R^{\frac{k}{k+1}}\right]x_{k+1}\\ &=(k+1)\sqrt[k+1]{x_1x_2\cdots x_{k+1}} \end{align*}

\100dpi \begin{align*} \frac{x_1+x_2+\cdots+x_{k+1}}{k+1}&\ge\frac{k\sqrt[k]{x_1x_2\cdots x_{k}}+x_{k+1}}{k+1}\\ &=\sqrt[k]{x_1x_2\cdots x_{k}}+\frac{x_{k+1}-\sqrt[k]{x_1x_2\cdots x_{k}}}{k+1}\\ &=\sqrt[k+1]{\left(\sqrt[k]{x_1x_2\cdots x_{k}}+\frac{x_{k+1}-\sqrt[k]{x_1x_2\cdots x_{k}}}{k+1}\right)^{k+1}}\\ &>\sqrt[k+1]{\left(\sqrt[k]{x_1x_2\cdots x_{k}}\right)^{k+1}+(k+1)\left(\sqrt[k]{x_1x_2\cdots x_{k}}\right)^{k}\left(\frac{x_{k+1}-\sqrt[k]{x_1x_2\cdots x_{k}}}{k+1}\right)}\\ &=\sqrt[k+1]{x_1x_2\cdots x_{k+1}} \end{align*}