设 $a,b, c$ 是给定复数，记 $|a+b|=m,|a-b|=n$．已知 $m n \neq 0$，求证 $\max \{|a c+b|,|a+b c|\} \geqslant \dfrac{m n}{\sqrt{m^{2}+n^{2}}}$
【难度】
【出处】
2007第22届CMO试题
【标注】
【答案】
略
【解析】
证法一
因为 $\max \{|a c+b|,|a+b c|\} \geqslant\dfrac{|b| \cdot|a c+b|+|a| \cdot|a+b c|}{|b|+|a|} \geqslant\dfrac{|b(a c+b)-a(a+b c)|}{|a|+|b|}=\dfrac{\left|b^{2}-a^{2}\right|}{|a|+|b|} \geqslant\dfrac{|b+a| \cdot|b-a|}{\sqrt{2\left(|a|^{2}+|b|^{2}\right)}}$ 又 $m^{2}+n^{2}=|a-b|^{2}+|a+b|^{2}=2\left(|a|^{2}+|b|^{2}\right)$ 所以 $\max \{|a c+b|,|a+b c|\} \geqslant \dfrac{m n}{\sqrt{m^{2}+n^{2}}}$
证法二
注意到 $a c+b=\dfrac{1+c}{2}(a+b)-\dfrac{1-c}{2}(a-b),a+b c=\dfrac{1+c}{2}(a+b)+\dfrac{1-c}{2}(a-b)$
令 $\alpha=\dfrac{1+c}{2}(a+b), \beta=\dfrac{1-c}{2}(a-b)$ 则 $\begin{aligned}|a c+b|^{2}+1 a+b\left.c\right|^{2}=&|\alpha-\beta|^{2}+|\alpha+\beta|^{2}= 2\left(|\left.\alpha\right|^{2}+|\beta|^{2}\right) \end{aligned}$
所以 $\begin{aligned}(\max \{|a c+b|,|a+b c|\})^{2} \geqslant &|\alpha|^{2}+|\beta|^{2}=\left|\dfrac{1+c}{2}\right|^{2} m^{2}+\left|\dfrac{1-c}{2}\right|^{2} n^{2} \end{aligned}$
因此只要证明 $\left|\dfrac{1+c}{2}\right|^{2} m^{2}+\left|\dfrac{1-c}{2}\right|^{2} n^{2} \geqslant \dfrac{m^{2} n^{2}}{m^{2}+n^{2}}$
等价变形为 $\left|\dfrac{1+c}{2}\right|^{2} m^{4}+\left|\dfrac{1-c}{2}\right|^{2} n^{4}+\left(\left|\dfrac{1+c}{2}\right|^{2}+\left|\dfrac{1-c}{2}\right|^{2}\right) m^{2} n^{2} \geqslant m^{2} n^{2}$ ①
事实上
$\left|\dfrac{1+c}{2}\right|^{2} m^{4}+\left|\dfrac{1-c}{2}\right|^{2} n^{4}+\left(\left|\dfrac{1+c}{2}\right|^{2}+\left|\dfrac{1-c}{2}\right|^{2}\right) m^{2} n^{2} \geqslant2\left|\dfrac{1+c}{2}\right|\left|\dfrac{1-c}{2}\right| m^{2} n^{2}+\left(\left|\dfrac{1+2 c+c^{2}}{4}\right|+\left|\dfrac{1-2 c+c^{2}}{4}\right|\right) m^{2} n^{2}\\=\left(\left|\dfrac{1-c^{2}}{2}\right|+\left|\dfrac{1+2 c+c^{2}}{4}\right|+\left|\dfrac{1-2 c+c^{2}}{4}\right|\right) m^{2} n^{2} \geqslant\left|\dfrac{1-c^{2}}{2}+\dfrac{1+2 c+c^{2}}{4}+\dfrac{1-2 c+c^{2}}{4}\right| m^{2} n^{2}=m^{2} n^{2}$
故 ① 得证．
证法三
由已知得
$m^{2}=|a+b|^{2}=(a+b)(\overline{a}+b)=(a+b)(\overline{a}+\overline{b})=
|a|^{2}+|b|^{2}+a \overline{b}+\overline{a} b$
$n^{2}=|a-b|^{2}=(a-b)(\overline{a-b})=(a-b)(\overline{a}-\overline{b})=|a|^{2}+|b|^{2}-a \overline{b}-\overline{a} b$
从而 $|a|^{2}+|b|^{2}=\dfrac{m^{2}+n^{2}}{2}, a \overline{b}+\overline{a} b=\dfrac{m^{2}-n^{2}}{2}$
令 $c=x+y i, x, y \in \mathbf{R}$，则
$|a c+b|^{2}+|a+b c|^{2}=$
$(a c+b)\left(\dfrac{a c+b}{a c+b}\right)+(a+b c)(\overline{a+b c})=$
$|a|^{2}|c|^{2}+|b|^{2}+a \overline{b} c+a b c+|a|^{2}+$
$|b|^{2}|c|^{2}+\overline{a} b c+a \overline{b c}=$
$\left(|c|^{2}+1\right)\left(|a|^{2}+|b| a^{2}\right)+(c+\overline{c})(a \overline{b}+\overline{a} b)=$
$\left(x^{2}+y^{2}+1\right) \dfrac{m^{2}+n^{2}}{2}+2 x \dfrac{m^{2}-n^{2}}{2} \geqslant$
$\dfrac{m^{2}+n^{2}}{2} x^{2}+\left(m^{2}-n^{2}\right) x+\dfrac{m^{2}+n^{2}}{2}=$
$\dfrac{m^{2}+n^{2}}{2}\left(x+\dfrac{m^{2}-n^{2}}{m^{2}+n^{2}}\right)^{2}-\dfrac{m^{2}+n^{2}}{2}\left(\dfrac{m^{2}-n^{2}}{m^{2}+n^{2}}\right)^{2}+\dfrac{m^{2}+n^{2}}{2} \geqslant$
$\dfrac{m^{2}+n^{2}}{2}-\dfrac{1}{2} \dfrac{\left(m^{2}-n^{2}\right)^{2}}{m^{2}+n^{2}}=\dfrac{2 m^{2} n^{2}}{m^{2}+n^{2}}$
所以 $(\max \{|a c+b|,|a+b c|\})^{2} \geqslant \dfrac{m^{2} n^{2}}{m^{2}+n^{2}}$ 即 $\max \{|a c+b|,|a+b c|\} \geqslant \dfrac{m n}{\sqrt{m^{2}+n^{2}}}$
因为 $\max \{|a c+b|,|a+b c|\} \geqslant\dfrac{|b| \cdot|a c+b|+|a| \cdot|a+b c|}{|b|+|a|} \geqslant\dfrac{|b(a c+b)-a(a+b c)|}{|a|+|b|}=\dfrac{\left|b^{2}-a^{2}\right|}{|a|+|b|} \geqslant\dfrac{|b+a| \cdot|b-a|}{\sqrt{2\left(|a|^{2}+|b|^{2}\right)}}$ 又 $m^{2}+n^{2}=|a-b|^{2}+|a+b|^{2}=2\left(|a|^{2}+|b|^{2}\right)$ 所以 $\max \{|a c+b|,|a+b c|\} \geqslant \dfrac{m n}{\sqrt{m^{2}+n^{2}}}$
证法二
注意到 $a c+b=\dfrac{1+c}{2}(a+b)-\dfrac{1-c}{2}(a-b),a+b c=\dfrac{1+c}{2}(a+b)+\dfrac{1-c}{2}(a-b)$
令 $\alpha=\dfrac{1+c}{2}(a+b), \beta=\dfrac{1-c}{2}(a-b)$ 则 $\begin{aligned}|a c+b|^{2}+1 a+b\left.c\right|^{2}=&|\alpha-\beta|^{2}+|\alpha+\beta|^{2}= 2\left(|\left.\alpha\right|^{2}+|\beta|^{2}\right) \end{aligned}$
所以 $\begin{aligned}(\max \{|a c+b|,|a+b c|\})^{2} \geqslant &|\alpha|^{2}+|\beta|^{2}=\left|\dfrac{1+c}{2}\right|^{2} m^{2}+\left|\dfrac{1-c}{2}\right|^{2} n^{2} \end{aligned}$
因此只要证明 $\left|\dfrac{1+c}{2}\right|^{2} m^{2}+\left|\dfrac{1-c}{2}\right|^{2} n^{2} \geqslant \dfrac{m^{2} n^{2}}{m^{2}+n^{2}}$
等价变形为 $\left|\dfrac{1+c}{2}\right|^{2} m^{4}+\left|\dfrac{1-c}{2}\right|^{2} n^{4}+\left(\left|\dfrac{1+c}{2}\right|^{2}+\left|\dfrac{1-c}{2}\right|^{2}\right) m^{2} n^{2} \geqslant m^{2} n^{2}$ ①
事实上
$\left|\dfrac{1+c}{2}\right|^{2} m^{4}+\left|\dfrac{1-c}{2}\right|^{2} n^{4}+\left(\left|\dfrac{1+c}{2}\right|^{2}+\left|\dfrac{1-c}{2}\right|^{2}\right) m^{2} n^{2} \geqslant2\left|\dfrac{1+c}{2}\right|\left|\dfrac{1-c}{2}\right| m^{2} n^{2}+\left(\left|\dfrac{1+2 c+c^{2}}{4}\right|+\left|\dfrac{1-2 c+c^{2}}{4}\right|\right) m^{2} n^{2}\\=\left(\left|\dfrac{1-c^{2}}{2}\right|+\left|\dfrac{1+2 c+c^{2}}{4}\right|+\left|\dfrac{1-2 c+c^{2}}{4}\right|\right) m^{2} n^{2} \geqslant\left|\dfrac{1-c^{2}}{2}+\dfrac{1+2 c+c^{2}}{4}+\dfrac{1-2 c+c^{2}}{4}\right| m^{2} n^{2}=m^{2} n^{2}$
故 ① 得证．
证法三
由已知得
$m^{2}=|a+b|^{2}=(a+b)(\overline{a}+b)=(a+b)(\overline{a}+\overline{b})=
|a|^{2}+|b|^{2}+a \overline{b}+\overline{a} b$
$n^{2}=|a-b|^{2}=(a-b)(\overline{a-b})=(a-b)(\overline{a}-\overline{b})=|a|^{2}+|b|^{2}-a \overline{b}-\overline{a} b$
从而 $|a|^{2}+|b|^{2}=\dfrac{m^{2}+n^{2}}{2}, a \overline{b}+\overline{a} b=\dfrac{m^{2}-n^{2}}{2}$
令 $c=x+y i, x, y \in \mathbf{R}$，则
$|a c+b|^{2}+|a+b c|^{2}=$
$(a c+b)\left(\dfrac{a c+b}{a c+b}\right)+(a+b c)(\overline{a+b c})=$
$|a|^{2}|c|^{2}+|b|^{2}+a \overline{b} c+a b c+|a|^{2}+$
$|b|^{2}|c|^{2}+\overline{a} b c+a \overline{b c}=$
$\left(|c|^{2}+1\right)\left(|a|^{2}+|b| a^{2}\right)+(c+\overline{c})(a \overline{b}+\overline{a} b)=$
$\left(x^{2}+y^{2}+1\right) \dfrac{m^{2}+n^{2}}{2}+2 x \dfrac{m^{2}-n^{2}}{2} \geqslant$
$\dfrac{m^{2}+n^{2}}{2} x^{2}+\left(m^{2}-n^{2}\right) x+\dfrac{m^{2}+n^{2}}{2}=$
$\dfrac{m^{2}+n^{2}}{2}\left(x+\dfrac{m^{2}-n^{2}}{m^{2}+n^{2}}\right)^{2}-\dfrac{m^{2}+n^{2}}{2}\left(\dfrac{m^{2}-n^{2}}{m^{2}+n^{2}}\right)^{2}+\dfrac{m^{2}+n^{2}}{2} \geqslant$
$\dfrac{m^{2}+n^{2}}{2}-\dfrac{1}{2} \dfrac{\left(m^{2}-n^{2}\right)^{2}}{m^{2}+n^{2}}=\dfrac{2 m^{2} n^{2}}{m^{2}+n^{2}}$
所以 $(\max \{|a c+b|,|a+b c|\})^{2} \geqslant \dfrac{m^{2} n^{2}}{m^{2}+n^{2}}$ 即 $\max \{|a c+b|,|a+b c|\} \geqslant \dfrac{m n}{\sqrt{m^{2}+n^{2}}}$
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