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广义平均不等式(默认数域为\ (\ mathbbr r \):

\(\forall a_i0\),\(r_1,r_2\neq 0\),\(r_1r_2\),全部。

\[{\sum_{i=1}^n}^{\frac{1}{r_1}}\frac{1}{n}a_i^{r_1}\le{\sum_{i=1}^n}^{\frac{1}{r_2}}\frac{1}{n}a_i^{r_2}

\]

先证明当\(n=2\)时的情况。设\(c=\frac{a_1}{a_2}\),则原公式等价于:

\[(\ frac { 1 c^{r_1}}{2})^{\frac{1}{r_1}}\le(\frac{1 c^{r_2}}{2})^\frac{1}{r_2}

\]

让我们考虑一下函数\(f\)

\[f(x)=(\ frac { 1 c^x}{2})^\frac{1}{x}

\]

请注意,此函数的域是\ (\ mathbr-\ {0 \} \)。然后考虑它的对数函数\(g\)

\[g(x)=\ ln f(x)=\ frac { 1 } { x } \ ln \ frac { 1 c^x}{2}

\]

只要证明\(g\)是单调的。取\(g\)的导数,有

\[g'(x)=-\frac{1}{x^2}\ln\frac{1 c^x}{2} { 1 } { x } \ cdot \ frac { 2 } { 1 c^x}\cdot\frac{c^x\ln c } { 2 } \ \

=-\frac{1}{x^2}\ln\frac{1 c^x}{2} \frac{1}{x}\cdot\frac{c^x\lnc } { 1 c^x}\\

=\ frac { 1 } { x }(-\ frac { 1 } { x } \ ln \ frac { 1 c^x}{2} \frac{c^x\ln c } { 1 c^x})\\

=\ frac { 1 } { x }(-\ ln(\ frac { 1 c^x}{2})^{\frac{1}{x}} \ c^{\frac{c^x}{1 c^x}})

\]

分别证明\(x0\)和\(x0\)的\(g'(x)\ge0\)。当\(x0\),\(\frac{1}{x}0\),只需证明

\[c^{\frac{c^x}{1 c^x}}\ge(\frac{1 c^x}{2})^{\frac{1}{x}}\\

\ IFF({c^{\frac{c^x}{1 c^x}}})^{x(1 c^x)}\ge{(\frac{1 c^x}{2})^{\frac{1}{x}}}^{x(1 c^x)} \

\ IFF({c^x})^{c^x}\ge(\frac{1 c^x}{2})^{1 c^x}\\

\]

设\ (t=c x \),则上述公式等价于\ (t t \ ge (\ frac {1t} {2}) {1t} \)。设\(h(t)=t \ ln t-(1t)\ ln(\ frac { 1t } { 2 })\)。只有\(h(t)\ge 0\)是必需的。求导数并得到它。

\[h '(t)=\ ln t 1-(\ ln \ frac { 1t } { 2 }(1t)\ cdot \ frac { 2 } { 1t } \ cdot \ frac { 1 } { 2 })\ \

=\ln t-\ln\frac{1 t}{2}\\

=\ln\frac{2t}{1 t}\\

=\ln(2-\frac{2}{1 t})

\]

不难发现\(h(t)\)在区间\((0,1)\)减小,在区间\([1,\infty)\)增大。还注意到\(h\)的定义域是\((0,\infty)\),所以\(h(1)=0\)是\(h\)的全局最小值。因此,\(c { \ frac { c x } { 1c x } } \ ge(\ frac { 1c x } { 2 }){ \ frac { 1 } { x } } \),即当\(x0\)

如果\(x0\)呢?此时,\(\frac{1}{x}0\),但在使用\(t\)改变元素之前,需要改变不等式左右两边的\ (x (1C x) \)次幂的不等号的符号;剩下的过程和上面类似。至此,我们已经证明了二元广义平均不等式的情况。

现在考虑扩展到\(n\)元。我们首先采用反向归纳法.证明了当\ (\ for all m \ in \ mathbb z,n=2 m \)时,广义平均不等式成立。

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