174. Some interesting properties of modulus (middle school).

by Virgil Nicula, Nov 21, 2010, 10:06 AM

$\left\{\begin{array}{cc}
1\blacktriangleright & |x|=\max\ \{x,-x\}\ \ \wedge\ \ |x+y|\le |x|+|y|\\\\
2\blacktriangleright & \left\{\begin{array}{c}
2\cdot\max\ \{x,y\}=x+y+|x-y|\\\\
2\cdot\min\ \{x,y\}=x+y-|x-y|\end{array}\right|\\\\
3\blacktriangleright  & |x|+|y|\le 1\ \iff\ |x+y|\le 1\ \wedge\ |x-y|\le 1\\\\
4\blacktriangleright & (\forall )\ \epsilon >0\ ,\ x\in (l-\epsilon , l+\epsilon )\ \iff\ |x-l|\ <\ \epsilon\\\\
5\blacktriangleright & x\in [a,b]\cup [b,a]\iff (x-a)(x-b)\le 0\iff \left|x-\frac {a+b}{2}\right|\le\frac {|a-b|}{2}\iff |x-\min \{a,b\}|+|x-\max \{a,b\}|=|a-b|\end{array}\right\|$ .

Remark. $|x+y|\le |x|+|y|\iff$ $(x+y)^2\le x^2+y^2+2|xy|\iff$ $xy\le |xy|$ , what is truly. Observe that $|x+y|=|x|+|y|\iff xy\ge 0$ .

Here is and a nice proof of the equivalence $\blacktriangleleft3\blacktriangleright\ :\ |x|+|y|\le 1\iff$ $x^2+y^2+2|xy|\le 1\iff$ $2\cdot\max\{xy,-xy\}=$

$2\cdot |xy|\ \le\ 1-x^2-y^2\iff$ $\left\|\begin{array}{ccc}
+2xy & \le & 1-x^2-y^2\\\\
-2xy & \le & 1-x^2-y^2\end{array}\right\|\iff$ $\left\|\begin{array}{ccc}
(x+y)^2 & \le & 1\\\\
(x-y)^2 & \le & 1\end{array}\right\|\iff$ $\left\|\begin{array}{ccc}
|x+y| & \le & 1\\\\
|x-y| & \le & 1\end{array}\right\|$ .


Proposed problem 1. Solve the inequation $ | x^{2}-12|x|+20|\le 9$ .

Proof. Observe that $x^2=|x|^2\implies  x^{2}-12|x|+20=(|x|-2)(|x|-10)$ . I"ll use the substitution $\boxed{\ |x|-2=t\ge -2\ }\ (*)$ .

Our inequality becomes $|t(t-8)|\le 9\iff$ $\left(t^2-8t-9\right)\left(t^2-8t+9\right)\le 0\iff$ $t\in\left[-1,4-\sqrt 7\right]\cup\left[4+\sqrt 7,9\right]\stackrel{(*)}{\iff}$

$|x|\in\left[1,6-\sqrt 7\right]\cup\left[6+\sqrt 7,11\right]\iff$ $\boxed{\boxed{\ x\in\left[-11,-6-\sqrt 7\right]\cup\left[-6+\sqrt 7,-1\right]\cup\left[1,6-\sqrt 7\right]\cup\left[6+\sqrt 7,11\right]\ }}$ .


Proposed problem 2. Solve the following system $\left\{\begin{array}{c}
a < x+c < b\\\
a < x-c < b\end{array}\right\|$ , where $a$ , $b$ , $c$ are real numbers.

Proof. $\left\|\begin{array}{c}
a< x+c< b\\\
a< x-c< b\end{array}\right\|\iff$ $\left\|\begin{array}{c}
a-c< x< b-c\\\
a+c< x< b+c\end{array}\right\|\iff$ $\max \{a-c,a+c\}<x<\min\{b-c,b+c\}\iff$

$a+\max \{-c,c\}<x<b+\min \{-c,c\}\iff$ $a+|c|<x<b-|c|\iff$ $x\in \left\{\begin{array}{ccc}
(a+|c|,b-|c|) & \mathrm{if} & |c|<\frac {b-a}{2}\\\\
\emptyset & \mathrm{if} & |c|\ge\frac {b-a}{2}\end{array}\right\|$ .



Proposed problem 3. Find the solution of the inequality $\left|\frac{3x+8}{3x^2+17x+24}\right|\ge 5$ .

Proof. Observe that $3x^2+17x+24=(x+3)(3x+8)$ and our inequality becomes $\left|\frac {3x+8}{(x+3)(3x+8)}\right|$ $\ge 5\iff\frac {1}{|x+3|}\ge 5\iff$

$|x+3|$ $\le\frac 15\iff$ $x\in\left[-3-\frac 15,-3+\frac 15\right]$ , where $x\not\in \left\{-3,-\frac 83\right\}$ . In conclusion, $x\in\left[-\frac {16}{5},-\frac {14}{5}\right]\ ,\ x\ne -3$ because $-\frac {16}{5}<-3<-\frac {14}{5}<-\frac 83$ .
This post has been edited 35 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:09 PM

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