Answer to Question #157324 in Calculus for Janhavi

Question

Prove that |x+y+z|<=|x|+|y|+|z|

Answer 1

$\text {let }a,b\in{R}$

$a*b \le|a|*|b|$

$(a+b)^2=a^2+b^2+2a*b\le{a^2+b^2+2|a|*|b|}$

$a^2+b^2+2|a|*|b|=|a|^2+|b|^2+2|a|*|b|=(|a|+|b|)^2$

$(a+b)^2\le(|a|+|b|)^2$

$\sqrt{(a+b)^2}\le\sqrt{(|a|+|b|)^2}$

$|a+b|\le|a|+|b|\ (1)$

$|x+y+z| \text{ let }a=x+y;b=z$

$\text{from inequality }(1)$

$|x+y+z|\le|x+y|+|z|\le|x|+|y|+|z|$

proved.