China Mathematical Olympiad

Proof I We have $$\begin{array}{rl}\max \{|ac+b|,|a+bc|\}& \ge \frac{|b|\cdot |ac+b|+|a|\cdot |a+bc|}{|b|+|a|}\\ & \ge \frac{|b(ac+b)-a(a+bc)|}{|a|+|b|}\\ & =\frac{|b^2-a^2|}{|a|+|b|}\\ & \ge \frac{|b+a|\cdot |b-a|}{\sqrt{2(|a{|}^2+|b{|}^2)}}.\end{array}$$ As $$m^2+n^2=|a-b{|}^2+|a+b{|}^2=2(|a{|}^2+|b{|}^2),$$ we get $$\max \{|ac+b|,|a+bc|\}\ge \frac{mn}{\sqrt{m^2+n^2}}.$$

Proof II Note that $$ac+b=\frac{1+c}{2}(a+b)-\frac{1-c}{2}(a-b)$$ and $$a+bc=\frac{1+c}{2}(a+b)+\frac{1-c}{2}(a-b).$$ Let $\alpha =\frac{1+c}{2}(a+b)$ and $\beta =\frac{1-c}{2}(a-b)$. Then $$\begin{array}{rl}|ac+b{|}^2+|a+bc{|}^2& =|\alpha -\beta {|}^2+|\alpha +\beta {|}^2\\ & =2(|\alpha {|}^2+|\beta {|}^2).\end{array}$$ Therefore $$\begin{array}{rl}(\max \{|ac+b|,|a+bc|\}{)}^2& \ge |\alpha {|}^2+|\beta {|}^2\\ & ={\left|\frac{1+c}{2}\right|}^2{m}^2+{\left|\frac{1-c}{2}\right|}^2{n}^2.\end{array}$$ Now we only need to prove that $${\left|\frac{1+c}{2}\right|}^2{m}^2+{\left|\frac{1-c}{2}\right|}^2{n}^2\ge \frac{m^2{n}^2}{m^2+n^2},$$ or equivalently $${\left|\frac{1+c}{2}\right|}^2{m}^4+{\left|\frac{1-c}{2}\right|}^2{n}^4+\left({\left|\frac{1+c}{2}\right|}^2+{\left|\frac{1-c}{2}\right|}^2\right)m^2{n}^2\ge m^2{n}^2.$$ We have $$\begin{array}{rl}& {\left|\frac{1+c}{2}\right|}^2{m}^4+{\left|\frac{1-c}{2}\right|}^2{n}^4+\left({\left|\frac{1+c}{2}\right|}^2+{\left|\frac{1-c}{2}\right|}^2\right)m^2{n}^2\\ \ge & 2\left|\frac{1+c}{2}\right|\left|\frac{1-c}{2}\right|m^2{n}^2+\left(\left|\frac{1+2c+c^2}{4}\right|+\left|\frac{1-2c+c^2}{4}\right|\right)m^2{n}^2\\ =& \left(\left|\frac{1-c^2}{2}\right|+\left|\frac{1+2c+c^2}{4}\right|+\left|\frac{1-2c+c^2}{4}\right|\right)m^2{n}^2\\ \ge & \left|\frac{1-c^2}{2}+\frac{1+2c+c^2}{4}+\frac{1-2c+c^2}{4}\right|m^2{n}^2\\ =& m^2{n}^2.\end{array}$$ This completes the proof.

Proof III Since $$\begin{array}{rl}{m}^2& =|a+b{|}^2=(a+b)(\overline{a}+\overline{b})=(a+b)(\bar{a}+\bar{b})\\ & =|a^2|+|b^2|+\overline{a\bar{b}}+\overline{ab},\\ n^2& =|a-b{|}^2=(a-b)(\overline{a}-\overline{b})=(a-b)(\bar{a}-\bar{b})\\ & =|a^2|+|b^2|-\overline{a\bar{b}}-\overline{ab},\end{array}$$ We get $$\begin{array}{rl}|a{|}^2+|b{|}^2& =\frac{m^2+n^2}{2},\\ \overline{ab}+\overline{ab}& =\frac{m^2-n^2}{2}.\end{array}$$ Let $c=x+yi$, $x,y\in \mathbb{R}$. Then $$\begin{array}{rl}& |ac+b{|}^2+|a+bc{|}^2\\ & =(ac+b)(\overline{ac}+\overline{b})+(a+bc)(\overline{a}+\overline{bc})\\ & =|a{|}^2|c{|}^2+|b{|}^2+\overline{a\bar{b}c}+\overline{a\bar{b}c}+|a{|}^2\\ & +|b{|}^2|c{|}^2+\overline{a\bar{b}c}+a\bar{b}\bar{c}\\ & =(|c{|}^2+1)(|a{|}^2+|b{|}^2)+(c+\bar{c})(\overline{a\bar{b}}+\overline{a\bar{b}})\\ & =(x^2+y^2+1)\frac{m^2+n^2}{2}+2x\frac{m^2-n^2}{2}\\ & \ge \frac{m^2+n^2}{2}{x}^2+(m^2-n^2)x+\frac{m^2+n^2}{2}\\ & =\frac{m^2+n^2}{2}{\left(x+\frac{m^2-n^2}{m^2+n^2}\right)}^2\\ & -\frac{m^2+n^2}{2}{\left(\frac{m^2-n^2}{m^2+n^2}\right)}^2+\frac{m^2+n^2}{2}\\ & \ge \frac{m^2+n^2}{2}-\frac{1}{2}\frac{(m^2-n^2{)}^2}{m^2+n^2}\\ & =\frac{2m^2{n}^2}{m^2+n^2}.\end{array}$$ That is $$(\max \{|ac+b|,|a+bc|\}{)}^2\ge \frac{m^2{n}^2}{m^2+n^2}.$$ Or $$\max \{|ac+b|,|a+bc|\}\ge \frac{mn}{\sqrt{m^2+n^2}}.$$