قدرمطلق

$\begin{array}{l}a\in ℝ\end{array}$

$if\left\{\begin{array}{l}x\ge 0\Rightarrow -x\le 0\Rightarrow \left|-x\right|=-\left(-x\right)=x\\ x<0\Rightarrow -x>0\Rightarrow \left|-x\right|=-a\end{array}\right.$

$\left\{\begin{array}{l}\left|-x\right|=x\\ \left|-x\right|=-x\end{array}〉\right.\left|-x\right|=\left|x\right|$

حالت اول-

فرض کنیم $\left|x\right|\le a$ باشد، ثابت می‌کنیم:

$-a\le x\le a$

بنابراین داریم:

$\left|x\right|\le a\Rightarrow {\left|x\right|}^2\le {\left(a\right)}^2\Rightarrow x^2\le a^2\Rightarrow x^2-a^2\le 0\Rightarrow \left(x-a\right)\left(x+a\right)\le 0$

نامساوی فوق را تعیین علامت می‌کنیم:

$\left(x-a\right)\left(x+a\right)=0\Rightarrow \left\{\begin{array}{l}x-a=0\Rightarrow x=a\\ x+a=0\Rightarrow x=-a\end{array}\right.$

$D=\left[-a,a\right]=\left\{x\left|-a\le x\le a\right.\right\}$

در فرایند فوق نامساوی قدرمطلقی $\left|x\right|\le a$ به نامساوی درجه دوم تبدیل و سپس تعیین علامت شده است. 

حالت دوم- 

فرض می‌کنیم $-a\le x\le a$ باشد، ثابت می‌کنیم:

$\left|x\right|\le a$

بنابراین داریم:

$-a\le x\le a\Rightarrow \left\{\begin{array}{l}-a\le x\Rightarrow x+a\ge 0\\ x\le a\Rightarrow x-a\le 0\end{array}〉\left(x+a\right)\left(x-a\right)\le 0\right.$

$\left(x+a\right)\left(x-a\right)\le 0\Rightarrow x^2-a^2\le 0\Rightarrow x^2\le a^2\Rightarrow \sqrt{x^2}\le \sqrt{a^2}\Rightarrow \left|x\right|\le \left|a\right|\stackrel{a\ge 0}{\to }\left|x\right|\le a$

فرض می‌کنیم $\left|x\right|\ge a$ باشد، ثابت می‌کنیم:

$\left\{\begin{array}{l}x\ge a\\ x\le -a\end{array}\right.$

بنابراین داریم:

$\left|x\right|\ge a\Rightarrow {\left|x\right|}^2\ge a^2\Rightarrow x^2\ge a^2\Rightarrow x^2-a^2\ge 0$

نامساوی فوق را تعیین علامت می‌کنیم:

$x^2-a^2=0\Rightarrow x^2=a^2\Rightarrow x=\pm a$

$D=\left\{x\left|x\le -a\right.\right.\vee \left.x\ge a\right\}$

با فرض این‌که $x$ و $y$ دو عدد حقیقی باشند، داریم: 

$\left|xy\right|=\sqrt{{\left(xy\right)}^2}=\sqrt{x^2{y}^2}=\sqrt{x^2}\sqrt{y^2}=\left|x\right|\left|y\right|$

$\left|\frac{x}{y}\right|=\sqrt{{\left(\frac{x}{y}\right)}^2}=\sqrt{\frac{x^2}{y^2}}=\frac{\sqrt{x^2}}{\sqrt{y^2}}=\frac{\left|x\right|}{\left|y\right|};y\ne 0$

$\left\{\begin{array}{l}x\ge 0\Rightarrow \left|x\right|=x\Rightarrow -\left|x\right|\le x\\ x<0\Rightarrow \left|x\right|=-x\Rightarrow x\le \left|x\right|\end{array}〉\right.-\left|x\right|\le x\le \left|x\right|$

روش اول:

$\left\{\begin{array}{l}-\left|x\right|\le x\le \left|x\right|\\ -\left|y\right|\le y\le \left|y\right|\end{array}〉\right.-\left(\left|x\right|+\left|y\right|\right)\le x+y\le \left(\left|x\right|+\left|y\right|\right)\Rightarrow \left|x+y\right|\le \left|x\right|+\left|y\right|$

روش دوم:

$\begin{array}{l}{\left|x+y\right|}^2=x^2+y^2+2xy\Rightarrow {\left|x+y\right|}^2\le x^2+y^2+2\left|x\right|\left|y\right|\Rightarrow {\left|x+y\right|}^2\le {\left(\left|x\right|+\left|y\right|\right)}^2\Rightarrow \left|x+y\right|\le \left|x\right|+\left|y\right|\end{array}$

$\begin{array}{l}\left|x+y\right|\le \left|x\right|+\left|y\right| ; x\to \left(x-y\right)\\ \\ \Rightarrow \left|\left(x-y\right)+y\right|\le \left|\left(x-y\right)\right|+\left|y\right|\\ \\ \Rightarrow \left|x\right|\le \left|x-y\right|+\left|y\right|\\ \\ \Rightarrow \left|x-y\right|\ge \left|x\right|-\left|y\right|\end{array}$

$\left|x+y\right|\le \left|x\right|+\left|y\right|\Rightarrow \left|x\right|=\left|\left(x+y\right)-y\right|\Rightarrow \left|x\right|\le \left|x+y\right|+\left|-y\right|\Rightarrow \left|x\right|\le \left|x+y\right|+\left|y\right|\Rightarrow \left|x+y\right|\ge \left|x\right|-\left|y\right|$

$\begin{array}{l}\left|x+y\right|\ge \left|y\right|-\left|x\right|\Rightarrow \left|x+y\right|\ge -\left(\left|x\right|-\left|y\right|\right)\Rightarrow \left|x\right|-\left|y\right|\ge -\left|x+y\right|\end{array}$

$\left\{\begin{array}{l}\left|x+y\right|\ge \left|x\right|-\left|y\right|\\ \left|x\right|-\left|y\right|\ge -\left|x+y\right|\end{array}\right.〉-\left|x+y\right|\le \left|x\right|-\left|y\right|\le \left|x+y\right|\Rightarrow \left|\left|x\right|-\left|y\right|\right|\le \left|x+y\right|$

$\left\{\begin{array}{l}x\ge 0\Rightarrow \left|x\right|=x\Rightarrow {\left|x\right|}^2=x^2\\ x<0\Rightarrow \left|x\right|=-x\Rightarrow {\left|x\right|}^2={\left(-x\right)}^2=x^2\end{array}〉\right.{\left|x\right|}^2=x^2$

$\begin{array}{l}\left|x\right|+\left|y\right|\ge \left|x+y\right|;\left\{\begin{array}{l}x\to x-z\\ y\to z-y\end{array}\right.\end{array}$

$\begin{array}{l}\Rightarrow \left|\left(x-z\right)\right|+\left|\left(z-y\right)\right|\ge \left|\left(x-z\right)+\left(z-y\right)\right|\end{array}$

$\Rightarrow \left|x-z\right|+\left|z-y\right|\ge \left|x-y\right|$

حالت اول:

$ifa>b\Rightarrow a-b>0\Rightarrow |a-b|=+(a-b)$

$\begin{array}{l}\max \left\{a,b\right\}=\frac{a+b}{2}+\frac{\left|a-b\right|}{2}=\frac{a+b}{2}+\frac{a-b}{2}=a\Rightarrow \max \left\{a,b\right\}=a\end{array}$

$\min \left\{a,b\right\}=\frac{a+b}{2}-\frac{\left|a-b\right|}{2}=\frac{a+b}{2}-\frac{a-b}{2}=b\Rightarrow \min \left\{a,b\right\}=b$

$\max \left\{a,b\right\}+\min \left\{a,b\right\}=a+b$

حالت دوم:

$ifa<b\Rightarrow a-b<0\Rightarrow \left|a-b\right|=-(a-b)$

$\begin{array}{l}\max \left\{a,b\right\}=\frac{a+b}{2}+\frac{\left|a-b\right|}{2}=\frac{a+b}{2}-\frac{a-b}{2}=b\Rightarrow \max \left\{a,b\right\}=b\end{array}$

$\min \left\{a,b\right\}=\frac{a+b}{2}-\frac{\left|a-b\right|}{2}=\frac{a+b}{2}+\frac{a-b}{2}=a\Rightarrow \min \left\{a,b\right\}=a$

$\max \left\{a,b\right\}+\min \left\{a,b\right\}=a+b$

نتیجه این قضیه به‌صورت زیر است:

$\begin{array}{l}\max \left\{a,0\right\}=\frac{a+\left|a\right|}{2}\\ \\ \min \left\{a,0\right\}=\frac{a-\left|a\right|}{2}\\ \\ \max \left\{a,b\right\}+\min \left\{a,b\right\}=a+b\end{array}$