算几不等式的证明(II)(Inequality of ari

作者: 分类: H嘉生活 发布于:2020-07-28 187次浏览 30条评论


连结:算几不等式的证明(I)

已知:\(a_1,a_2,…,a_n\) 为正数或零。

求证:

\(\displaystyle\frac{{{a_1} + {a_2} + … + {a_n}}}{n} \ge \sqrt[n]{{{a_1}{a_2}…{a_n}}}\),“\(=\)” 成立时若且唯若,\(a_1=a_2=…=a_n\)。

(其中 \(\displaystyle\frac{{{a_1} + {a_2} + … + {a_n}}}{n}\) 称为算术平均数,\(\sqrt[n]{{{a_1}{a_2}…{a_n}}}\) 称为几何平均数)。

除了常见的倒回的证明外(先证 \(n=2=2^1\) 再推论 \(n=4=2^2\),然后反推 \(n=3\),接着证明 \(n=8=2^3\),然后反推 \(n=5,6,7\) 等),在此处笔者再介绍两种直接证明方法。

第一种代换法:

\(n=2\),证明 \(\displaystyle\frac{{{a_1} + {a_2}}}{2} \ge \sqrt {{a_1}{a_2}}\) (不再赘述)

\(n=3\),证明 \(\displaystyle\frac{{{a_1} + {a_2} + {a_3}}}{3} \ge \sqrt[3]{{{a_1}{a_2}{a_3}}}\),

可令 \(\displaystyle{a_3} = \frac{{{a_3} + \sqrt[3]{{{a_1}{a_2}{a_3}}}}}{2}\),利用两数的算几不等式性质,即

\(\begin{array}{ll}\displaystyle\frac{\displaystyle{\frac{{{a_1} + {a_2}}}{2} + \frac{{{a_3} + \sqrt[3]{{{a_1}{a_2}{a_3}}}}}{2}}}{2}&\ge \sqrt {\left( {\displaystyle\frac{{{a_1} + {a_2}}}{2}} \right)\left( {\frac{{{a_3} + \sqrt[3]{{{a_1}{a_2}{a_3}}}}}{2}} \right)}\\&\ge \sqrt {\sqrt {{a_1}{a_2}} \sqrt {{a_3}\sqrt[3]{{{a_1}{a_2}{a_3}}}} }\\&= \sqrt[3]{{{a_1}{a_2}{a_3}}}\end{array}\)

左右两式整理得 \(\displaystyle\frac{{{a_1} + {a_2} + {a_3}}}{2} + \frac{{\sqrt[3]{{{a_1}{a_2}{a_3}}}}}{2} \ge 2\sqrt[3]{{{a_1}{a_2}{a_3}}}\),

移项得 \(\displaystyle\frac{{{a_1} + {a_2} + {a_3}}}{3} \ge \sqrt[3]{{{a_1}{a_2}{a_3}}}\),得证,等号成立的地方就请读者自行检验。

这个神乎其技的证明方式令人眼花了乱,姑且称之为「代换法」,所以,我们再试一次,

\(n=4\),证明 \(\displaystyle\frac{{{a_1} + {a_2} + {a_3} + {a_4}}}{4} \ge \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}\),利用两数及三数的算几不等式性质,

\(\displaystyle\frac{{\displaystyle\frac{{{a_1} + {a_2} + {a_3}}}{3} + \frac{{{a_4} + \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}} + \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}}}{3}}}{2}\\\ge\displaystyle \sqrt {\left( {\frac{{{a_1} + {a_2} + {a_3}}}{3}} \right)\left( {\frac{{{a_4} + \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}} + \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}}}{3}} \right)}\\\ge \sqrt {\sqrt[3]{{{a_1}{a_2}{a_3}}}\sqrt[3]{{{a_4}{{\left( {\sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}} \right)}^2}}}}\\= \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}\)

,左右两式整理得 \(\displaystyle\frac{{{a_1} + {a_2} + {a_3} + {a_4} + 2\sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}}}{3} \ge 2\sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}\),

移项得 \(\displaystyle\frac{{{a_1} + {a_2} + {a_3} + {a_4}}}{4} \ge \sqrt[4]{{{a_1}{a_2}{a_3}{a_4}}}\),

有了这两次的经验后,最后我们根据数学归纳法将 \(n=k\) 扩及到 \(n=k+1\),

\(\displaystyle\frac{\displaystyle{\frac{{{a_1} + … + {a_k}}}{k} + \frac{{{a_{k + 1}} + \sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}} + … + \sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}}}}{k}}}{2}\\\ge\sqrt {\left(\displaystyle{\frac{{{a_1} + … + {a_k}}}{k}} \right)\left( {\frac{{{a_{k + 1}} + \sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}} + … + \sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}}}}{k}} \right)}\\\ge \sqrt {\sqrt[k]{{{a_1}{a_2}…{a_k}}}\sqrt[k]{{{a_{k + 1}}{{\left( {\sqrt[{k + 1}]{{{a_1}{a_2}….{a_{k + 1}}}}} \right)}^{k – 1}}}}}\\= \sqrt {{a_1}^{\frac{1}{k}(1 + \frac{{k – 1}}{{k + 1}})}{a_2}^{\frac{1}{k}(1 + \frac{{k – 1}}{{k + 1}})}…{a_k}^{\frac{1}{k}(1 + \frac{{k – 1}}{{k + 1}})}{a_{k + 1}}^{\frac{1}{k}(1 + \frac{{k – 1}}{{k + 1}})}}\\= \sqrt[{k + 1}]{{{a_1}{a_2}…{a_k}{a_{k + 1}}}}\)

,左右两式整理得 \(\displaystyle\frac{{{a_1} + {a_2} + … + {a_{k + 1}} + \left( {k – 1} \right)\sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}}}}{k} \ge 2\sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}}\),

移项得 \({a_1} + {a_2} + … + {a_{k + 1}} \ge \left( {k + 1} \right)\sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}}\),

最后,\(\displaystyle\frac{{{a_1} + {a_2} + … + {a_{k + 1}}}}{{k + 1}} \ge \sqrt[{k + 1}]{{{a_1}{a_2}…{a_{k + 1}}}}\),得证。

这个证明巧妙地使用代换方法(substitution method),十分令人激赏。

第二种微分法:

我们可以使用微分方法,

首先考虑函数 \(\displaystyle{f}\left( x \right) = \frac{{{a_1} + {a_2} + … + {a_n} + x}}{{n + 1}} – {\left( {{a_1}{a_2}…{a_n}x} \right)^{\frac{1}{{n + 1}}}},x > 0\),

其中 \(a_1,a_2,…,a_n\) 为正数或零,算几不等式的等价证明为 \(\forall x>0,~f(x)\ge 0\),

因此,先将函数 \(f(x)\) 微分,得 \(\displaystyle{f'(x)}= \frac{1}{{n + 1}} – {\left( {{a_1}{a_2}…{a_n}} \right)^{\frac{1}{{n + 1}}}}\frac{1}{{n + 1}}{x^{\frac{1}{{n + 1}} – 1}}\),

考虑 \(f'(t)=0\),得 \(1 = {\left( {{a_1}{a_2}…{a_n}} \right)^{\frac{1}{{n + 1}}}}{t^{\frac{{ – n}}{{n + 1}}}}\),即 \({t^{\frac{n}{{n + 1}}}} = {\left( {{a_1}{a_2}…{a_n}} \right)^{\frac{1}{{n + 1}}}}\),

最后得 \(t = {\left( {{a_1}{a_2}..{a_n}} \right)^{\frac{1}{n}}}\),此为 \(a_1,a_2,…,a_n\) 的几何平均数。

再将 \(f'(x)\) 微分,得

\(\begin{array}{ll}\displaystyle f”(x) &=\displaystyle- {\left( {{a_1}{a_2}…{a_n}} \right)^{\frac{1}{{n + 1}}}}\frac{1}{{n + 1}}\left( {\frac{{ – n}}{{n + 1}}} \right){x^{\frac{{ – n}}{{n + 1}} – 1}} \\&=\displaystyle {\left( {{a_1}{a_2}…{a_n}} \right)^{\frac{1}{{n + 1}}}}\frac{n}{{{{(n + 1)}^2}}}{x^{\frac{{ – 2n – 1}}{{n + 1}}}} > 0,\forall x > 0\end{array}\)

将 \(x=t\) 代入,则 \(f”(t)>0\),可见 \(f(t)\) 为相对极小值,

\(\begin{array}{ll}f\left( t \right)&=\displaystyle \frac{{{a_1} + {a_2} + … + {a_n} + {{\left( {{a_1}{a_2}…{a_n}} \right)}^{\frac{1}{n}}}}}{{n + 1}} – {\left[ {{a_1}{a_2}…{a_n}{{({a_1}{a_2}…{a_n})}^{\frac{1}{n}}}} \right]^{\frac{1}{{n + 1}}}}\\&=\displaystyle \frac{{{a_1} + {a_2} + … + {a_n}}}{{n + 1}} + \frac{{{{({a_1}{a_2}…{a_n})}^{\frac{1}{n}}}}}{{n + 1}} – {\left[ {{{\left( {{a_1}{a_2}…{a_n}} \right)}^{1 + \frac{1}{n}}}} \right]^{\frac{1}{{n + 1}}}}\\&=\displaystyle\frac{{{a_1} + {a_2} + … + {a_n}}}{{n + 1}} + \frac{{{{({a_1}{a_2}…{a_n})}^{\frac{1}{n}}}}}{{n + 1}} – {\left( {{a_1}{a_2}…{a_n}} \right)^{\frac{1}{n}}}\\&=\displaystyle\frac{n}{{n + 1}}\left[ {\frac{{{a_1} + {a_2} + … + {a_n}}}{n} – {{\left( {{a_1}{a_2}…{a_n}} \right)}^{\frac{1}{n}}}} \right] \ge 0\end{array}\)

这个式子成立是根据数学归纳法前 \(n\) 的条件,且等号成立于 \(a_1=a_2=…=a_n\),

而 \(t = {\left( {{a_1}{a_2}..{a_n}} \right)^{\frac{1}{n}}} = {a_1} = {a_2} = … = {a_n}\),

因此,除非 \(a_1=a_2=…=a_n=x\),否则 \(f(x)>0\),

即 \(\displaystyle\frac{{{a_1} + {a_2} + … + {a_n} + x}}{{n + 1}} > {\left( {{a_1}{a_2}…{a_n}x} \right)^{\frac{1}{{n + 1}}}}\),得证。

最后,练习一下国立台中一中99年数学资赋优异鉴定试题第九题:

已知 \(-1\le x \le \frac{3}{2}\),则当 \(x=k\) 时,\(f(x)=(x+1)^3(3-2x)^2\) 有最大值 \(m\),

试求 \((k,m)\) 数对?

参考解法:利用五数的算几不等式,

\(\displaystyle\frac{\displaystyle{\frac{4}{3}(x + 1) + \frac{4}{3}(x + 1) + \frac{4}{3}(x + 1) + \left( {3 – 2x} \right) + \left( {3 – 2x} \right)}}{5} \ge \sqrt[5]{{{{\left[ {\frac{4}{3}(x + 1)} \right]}^3}{{\left( {3 – 2x} \right)}^2}}}\)

整理得 \(\displaystyle{2^5} \le \frac{{64}}{{27}}{(x + 1)^3}{(3 – 2x)^2}\),移项得 \(\displaystyle{f\left( x \right)}= {\left( {x + 1} \right)^3}{(3 – 2x)^2} \le \frac{{27}}{2}\),

所以,\(\displaystyle m=\frac{27}{2}\),等号成立时,\(\displaystyle\frac{4}{3}(x + 1) = 3 – 2x,\therefore x = k = \frac{1}{2}\),

得 \(\displaystyle\left( {k,m} \right) = \left( {\frac{1}{2},\frac{{27}}{2}} \right)\)。

连结: 算几不等式的证明(III)

参考资料:

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