Question

prove that|a+b|>=||a|+|b||​

Answer 1

$\displaystyle\longrightarrow\bf{|a+b|}=\sf{\sqrt{a^2+b^2+2ab\cos\theta}}$

$\displaystyle\longrightarrow\bf{|a|+|b|}=\sf{\sqrt{a^2+b^2+2ab}}$

Assume that,

$\displaystyle\longrightarrow\bf{|a+b|\geq|a|+|b|}\quad\quad\sf{\dots(1)}$

$\displaystyle\longrightarrow\sf{\sqrt{a^2+b^2+2ab\cos\theta}\geq\sqrt{a^2+b^2+2ab}}$

$\displaystyle\longrightarrow\sf{a^2+b^2+2ab\cos\theta\geq a^2+b^2+2ab}$

$\displaystyle\longrightarrow\sf{\cos\theta\geq1}$

But $\displaystyle\sf{-1\leq\cos\theta\leq1.}$

Hence our assumption is contradicted and thus disproved.

Besides,

$\displaystyle\longrightarrow\bf{|a+b|\leq|a|+|b|}$

QED