45412 三角形結構中的一個證題系統

### 一、三角形中的一個證題系統

\begin{align*} &\hskip -30pt s\!=\!x+y+z,\ \triangle\!=\!\sqrt{xyz(x+y+z)},\ r\!=\!\sqrt{\dfrac{xyz}{x+y+z}},\ R\!=\!\frac{(x+y)(y+z)(z+x)}{4\sqrt{xyz(x+y+z)}},\\ h_a=&\dfrac{2\sqrt{xyz(x+y+z)}}{y+z},\quad h_b=\dfrac{2\sqrt{xyz(x+y+z)}}{z+x},\quad h_c=\dfrac{2\sqrt{xyz(x+y+z)}}{x+y},\\ m_a=&\dfrac 12\sqrt{4x^2\!+\!y^2\!+\!z^2\!+\!4xy\!+\!4zx\!-\!2yz},\ m_b=\dfrac 12\sqrt{4y^2\!+\!x^2\!+\!z^2\!+\!4yz\!+\!4yx\!-\!2xz},\\ m_c=&\dfrac 12\sqrt{4z^2\!+\!x^2\!+\!y^2\!+\!4zx\!+\!4zy\!-\!2xy},\\ w_a=&\dfrac {2\sqrt{x(x+y)(x+z)(x+y+z)}}{2x+y+z},\qquad w_b=\dfrac {2\sqrt{y(y+z)(y+x)(x+y+z)}}{2y+x+z},\\ w_c=&\dfrac {2\sqrt{z(z+x)(z+y)(x+y+z)}}{2z+x+y},\\ r_a=&\dfrac{\sqrt{xyz(x+y+z)}}{x},\qquad r_b=\dfrac{\sqrt{xyz(x+y+z)}}{y},\qquad r_c=\dfrac{\sqrt{xyz(x+y+z)}}{z},\\ \sin A\!=&\dfrac{2\sqrt{xyz(x+y+z)}}{(x+y)(x+z)},\ \sin B\!=\!\dfrac{2\sqrt{xyz(x+y+z)}}{(y+x)(y+z)},\ \sin C\!=\!\dfrac{2\sqrt{xyz(x+y+z)}}{(z+x)(z+y)},\\ \cos A\!=&\dfrac{x(x+y+z)-yz}{(x+y)(x+z)},\ \ \cos B\!=\!\dfrac{y(x+y+z)-xz}{(y+x)(y+z)},\ \ \cos C\!=\!\dfrac{z(x+y+z)-xy}{(z+x)(z+y)},\\ \tan A\!=&\dfrac{2\sqrt{xyz(x+y+z)}}{x(x+y+z)-yz},\ \tan B\!=\!\dfrac{2\sqrt{xyz(x+y+z)}}{y(x+y+z)-xz},\ \tan C\!=\!\dfrac{2\sqrt{xyz(x+y+z)}}{z(x+y+z)-xy},\\ \sin \frac A2\!=&\sqrt{\dfrac{yz}{(x+y)(x+z)}},\ \quad \sin \frac B2\!=\!\sqrt{\dfrac{xz}{(y+x)(y+z)}},\quad \sin \frac C2\!=\!\sqrt{\dfrac{xy}{(z+x)(z+y)}},\\ \cos \frac A2\!=&\sqrt{\dfrac{x(x+y+z)}{(x+y)(x+z)}},\ \quad \cos \frac B2\!=\!\sqrt{\dfrac{y(x+y+z)}{(y+x)(y+z)}},\quad \cos \frac C2\!=\!\sqrt{\dfrac{z(x+y+z)}{(z+x)(z+y)}},\\ \tan \frac A2\!=&\sqrt{\dfrac{yz}{x(x+y+z)}},\qquad \ \tan \frac B2\!=\!\sqrt{\dfrac{xz}{y(x+y+z)}},\quad\ \tan \frac C2\!=\!\sqrt{\dfrac{xy}{z(x+y+z)}}. \end{align*}

### 二、應用舉例

\begin{align*} &\hskip -20pt a^2+b^2+c^2\ge 4\sqrt{3}\Delta +(a-b)^2+(b-c)^2+(c-a)^2\\ \Leftrightarrow\ &(x\!+\!y)^2\!+\!(y\!+\!z)^2\!+\!(z\!+\!x)^2\ge 4\sqrt{3xyz(x\!+\!y\!+\!z)}\!+\!(y\!-\!x)^2\!+\!(z\!-\!y)^2\!+\!(x\!-\!z)^2\\ \Leftrightarrow\ &4(xy+yz+zx)\ge 4\sqrt{3xyz(x+y+z)}\\ \Leftrightarrow\ &(xy+yz+zx)^2\ge {3xyz(x+y+z)}\\ \Leftrightarrow\ &x^2y^2+y^2z^2+z^2x^2\ge x^2yz+y^2zx+z^2xy\\ \Leftrightarrow\ &\dfrac 12[(xy-yz)^2+(yz-zx)^2+(zx-xy)^2]\ge 0. \end{align*}

\begin{align*} &\hskip -20pt a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc\\ \Leftrightarrow\ &2(y+z)^2x+2(z+x)^2y+2(x+y)^2z\le 3(x+y)(y+z)(z+x)\\ \Leftrightarrow\ &x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\ge 6xyz, \end{align*}

\begin{align*} &\hskip -20pt a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\\ =&\,(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)+(x+y)^2(y+z)(x-z)\\ =&\,2(x^3z+y^3x+z^3y-x^2yz-y^2zx-z^2xy)\\ =&\,2xyz\Big[\Big(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\Big)-(x+y+z)\Big]\\ =&\,2xyz\Big[\frac{(x-y)^2}{y}+\frac{(y-z)^2}{z}+\frac{(z-x)^2}{x}\Big]\\ =&\,\frac{2xyz}{x+y+z}(x+y+z)\Big[\frac{(x-y)^2}{y}+\frac{(y-z)^2}{z}+\frac{(z-x)^2}{x}\Big]\\ \ge&\,\frac{2xyz}{x+y+z}(|x-y|+|y-z|+|z-x|)^2\\ \ge&\,\frac{2xyz}{x+y+z}(|x-y|+|(y-z)+(z-x)|)^2\\ =&\,\frac{8xyz}{x+y+z}(x-y)^2 =8r^2(a-b)^2 \end{align*}

\begin{align*} & a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 8r^2(b-c)^2;\\ & a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 8r^2(c-a)^2. \end{align*}

\begin{align*} &\hskip -20pt \dfrac{1}{\tan^3\dfrac A2}+\dfrac{1}{\tan^3\dfrac B2}+\dfrac{1}{\tan^3\dfrac C2}\ge 9\sqrt{3}\\ \Leftrightarrow\quad & \dfrac{1}{\Big(\sqrt{\dfrac{yz}{x(x+y+z)}}\Big)^3}+\dfrac{1}{\Big(\sqrt{\dfrac{zx}{y(x+y+z)}}\Big)^3} +\dfrac{1}{\Big(\sqrt{\dfrac{xy}{z(x+y+z)}}\Big)^3}\ge 9\sqrt{3}\\ \Leftrightarrow\quad & x^3+y^3+z^3\ge 9\sqrt{3}\Big(\sqrt{\dfrac{xyz}{x+y+z}}\Big)^3, \end{align*}

\begin{align*} R=\ &\dfrac{(x+y)(y+z)(z+x)}{4\sqrt{xyz(x+y+z)}}\ge \frac{2\sqrt{xy}\cdot 2\sqrt{yz}\cdot 2\sqrt{zx}}{4\sqrt{xyz(x+y+z)}}=\frac{8xyz}{4\sqrt{xyz(x+y+z)}}\\ =\ &2\sqrt{\dfrac{xyz}{x+y+z}}=2r, \end{align*}

\begin{align*} &\hskip -20pt \dfrac{a^2}{r_b^2+r_c^2}+\dfrac{b^2}{r_c^2+r_a^2}+\dfrac{c^2}{r_a^2+r_b^2}\ge 2\\ \Leftrightarrow\quad & \dfrac{(y+z)^2}{\dfrac{xyz(x+y+z)}{y^2}+\dfrac{xyz(x+y+z)}{z^2}}+\dfrac{(z+x)^2}{\dfrac{xyz(x+y+z)}{z^2}+\dfrac{xyz(x+y+z)}{x^2}}\\ &\qquad +\dfrac{(x+y)^2}{\dfrac{xyz(x+y+z)}{x^2}+\dfrac{xyz(x+y+z)}{y^2}}\ge 2\\ \Leftrightarrow\quad & \frac{x^2y^2(x+y)^2}{x^2+y^2}+\frac{y^2z^2(y+z)^2}{y^2+z^2}+\frac{z^2x^2(z+x)^2}{z^2+x^2}\ge 2xyz(x+y+z),\\ &\hskip -20pt \because\ (x+y)^2(x^2y^2+y^2z^2+z^2x^2)-2xyz(x+y+z)(x^2+y^2)\\ =\ &(x+y)^2z^2(x^2+y^2)+(x+y)^2x^2y^2-2xyz(x+y+z)(x^2+y^2)\\ =\ &(x^2+y^2)^2z^2+(x+y)^2x^2y^2-2xyz(x+y)(x^2+y^2)\\ =\ &[(x^2+y^2)^2z]^2-2(x^2+y^2)z\cdot xy(x+y)+[xy(x+y)]^2\\ =\ &[(x^2+y^2)z-xy(x+y)]^2\ge 0\\ &\hskip -20pt \therefore \ (x+y)^2(x^2y^2+y^2z^2+z^2x^2)\ge 2xyz(x+y+z)(x^2+y^2),\\ &\hskip -20pt \therefore \ \frac{x^2y^2(x+y)^2}{x^2+y^2}\ge \frac{2xyz(x+y+z)x^2y^2}{x^2y^2+y^2z^2+z^2x^2}\\ \hbox{同理可證：} &\frac{y^2z^2(y+z)^2}{y^2+z^2}\ge \frac{2xyz(x+y+z)y^2z^2}{x^2y^2+y^2z^2+z^2x^2};\\ &\frac{z^2x^2(z+x)^2}{z^2+x^2}\ge \frac{2xyz(x+y+z)z^2x^2}{x^2y^2+y^2z^2+z^2x^2}.\\ \hbox{將以上三式兩邊分別相加, 可得} &\frac{x^2y^2(x+y)^2}{x^2+y^2}+\frac{y^2z^2(y+z)^2}{y^2+z^2}+\frac{z^2x^2(z+x)^2}{z^2+x^2}\ge 2xyz(x+y+z), \end{align*}

\begin{align*} &\hskip -20pt \Big(\dfrac{h_a}{r_a}\Big)^2 + \Big(\dfrac{h_b}{r_b}\Big)^2 + \Big(\dfrac{h_c}{r_c}\Big)^2 \ge 4\Big(\sin^2\dfrac A2+\sin^2\dfrac B2+\sin^2\dfrac C2\Big)\\ \Leftrightarrow\quad & \Big(\frac{2x}{y+z}\Big)^2+\Big(\frac{2y}{z+x}\Big)^2+\Big(\frac{2x}{x+y}\Big)^2\\ &\quad \ge 4\Big[\frac{yz}{(x\!+\!y)(x\!+\!z)}+\frac{zx}{(y\!+\!x)(y\!+\!z)}+\frac{xy}{(z\!+\!x)(z\!+\!y)}\Big]\\ \Leftrightarrow\quad & \Big[\Big(\frac{x}{y+z}\Big)^2+\Big(\frac{y}{z+x}\Big)^2+\Big(\frac{x}{x+y}\Big)^2\Big]\\ &\quad \ge \Big[\frac{yz}{(x\!+\!y)(x\!+\!z)}+\frac{zx}{(y\!+\!x)(y\!+\!z)}+\frac{xy}{(z\!+\!x)(z\!+\!y)}\Big]\\ \Leftrightarrow\quad & \Big(\frac x{y+z}\Big)^2+\Big(\frac y{x+z}\Big)^2+\Big(\frac y{x+z}\Big)^2+\Big(\frac z{x+y}\Big)^2+\Big(\frac z{x+y}\Big)^2 +\Big(\frac x{y+z}\Big)^2\\ &\ \ge 2\Big[\Big(\frac x{y+z}\Big)\cdot\Big(\frac y{x+z}\Big)+\Big(\frac y{x+z}\Big)\cdot \Big(\frac z{x+y}\Big)+\Big(\frac z{x+y}\Big)\cdot\Big(\frac x{y+z}\Big) \Big]\\ \Leftrightarrow\quad & \Big[\Big(\frac x{y+z}-\frac y{x+z}\Big)^2+\Big(\frac y{x+z}-\frac z{x+y}\Big)^2+\Big(\frac z{x+y}-\frac x{y+z}\Big)^2\Big]\ge 0. \end{align*}

\begin{align*} &\hskip -20pt \frac x{y+z}+\frac y{z+x}+\frac z{x+y}+6\Big(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\Big)\\ \ge\ &\frac 13(x+y+z)\Big(\frac 1{y+z}+\frac 1{z+x}+\frac 1{x+y}\Big)\\ &\ +6\times \frac 13(x+y+z) \Big(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\Big)\\ =\ &\frac 13(x+y+z)\Big[\Big(\frac {1^2}{y+z}+\frac {1^2}{z+x}+\frac {1^2}{x+y}\Big)\\ &\ +6\times \Big(\frac{1^2}{2x+y+z}+\frac{1^2}{x+2y+z}+\frac{1^2}{x+y+2z}\Big)\Big]\\ \ge\ &\frac 13(x+y+z)\Big[\frac {(1+1+1)^2}{y\!+\!z\!+\!z\!+\!x\!+\!x\!+\!y}+6\Big(\frac{(1+1+1)^2}{2x\!+\!y\!+\!z\!+\!x\!+\!2y\!+\!z\!+\!x\!+\!y\!+\!2z}\Big)\Big]\\ =\ &6. \end{align*}

\begin{align*} &\hskip -20pt ar_a+br_b+cr_c\ge \sqrt{bc}r_a+\sqrt{ca}r_b+\sqrt{ab}r_c\\ \Leftrightarrow\ & \frac{y\!+\!z}x\!+\!\frac{z\!+\!x}y\!+\!\frac {x\!+\!y}z\ge \frac{\sqrt{(z\!+\!x)(x\!+\!y)}}x\!+\!\frac{\sqrt{(x\!+\!y)(y\!+\!z)}}y\!+\!\frac {\sqrt{(y\!+\!z)(z\!+\!x)}}z\\ \end{align*}

\begin{align*} &\hskip -20pt \frac{\sqrt{(z+x)(x+y)}}x+\frac{\sqrt{(x+y)(y+z)}}y+\frac {\sqrt{(y+z)(z+x)}}z\\ \le\ & \frac{2x+y+z}{2x}+\frac{x+2y+z}{2y}+\frac{x+y+2z}{2z}, \end{align*}

### 參考文獻

---本文作者範花妹, 秦慶雄任教中國雲南省大理州漾濞縣第一中學(高中部)---