Question Number 164622 by mathocean1 last updated on 19/Jan/22
$${Show}\:{that}\:\forall\:{a},\:{b}\:\in\:\mathbb{R}, \\ $$$$\mathrm{1}.\:\mid\mid{x}\mid−\mid{y}\mid\mid\leqslant\mid{x}−{y}\mid \\ $$$$\mathrm{2}.\:\mathrm{1}+\mid{xy}−\mathrm{1}\mid\leqslant\left(\mathrm{1}+\mid{x}−\mathrm{1}\mid\right)\left(\mathrm{1}+\mid{y}−\mathrm{1}\mid\right). \\ $$
Answered by hmrsh last updated on 19/Jan/22
$$ \\ $$$$\mathrm{Answer}\:\mathrm{of}\:\mathrm{part}\left(\mathrm{1}\right): \\ $$$$\ast\:\mid{x}\mid\:\leqslant\:{A}\:\leftrightarrow\:−{A}\:\leqslant\:{x}\:\leqslant\:{A} \\ $$$$\ast\ast\:\mid{x}\:+\:{y}\mid\:\leqslant\:\mid{x}\mid\:+\:\mid{y}\mid\:\:\:\:\:\:\left(\mathrm{triangle}\:\mathrm{inequality}\right) \\ $$$$ \\ $$$$\mid{x}\mid\:=\:\mid\left({x}\:−\:{y}\right)\:+\:{y}\mid \\ $$$$\overset{\ast\ast} {\rightarrow}\:\mid{x}\mid\:=\:\mid\left({x}\:−\:{y}\right)\:+\:{y}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid\:+\:\mid{y}\mid\:\:\:\left(#\right) \\ $$$$\mathrm{also}: \\ $$$$\mid{y}\mid\:=\:\mid\left({y}−{x}\right)+{x}\mid\:\leqslant\:\mid{y}−{x}\mid+\mid{x}\mid=\mid{x}−{y}\mid+\mid{x}\mid \\ $$$$\rightarrow\:\mid{y}\mid\:−\mid{x}\:−\:{y}\mid\:\leqslant\:\mid{x}\mid \\ $$$$ \\ $$$$\overset{\left(#\right)} {\rightarrow}\:\mid{y}\mid\:−\mid{x}\:−\:{y}\mid\:\leqslant\:\mid{x}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid\:+\:\mid{y}\mid \\ $$$$\rightarrow\:−\mid{x}\:−\:{y}\mid\:\leqslant\:\mid{x}\mid\:−\:\mid{y}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid \\ $$$$\overset{\ast} {\rightarrow}\:\mid\mid{x}\mid\:−\:\mid{y}\mid\mid\:\leqslant\:\mid{x}\:−\:{y}\mid \\ $$
Commented by mathocean1 last updated on 19/Jan/22
$${thanks}. \\ $$
Answered by hmrsh last updated on 20/Jan/22
$$\mathrm{Answer}\:\mathrm{of}\:\mathrm{part}\:\mathrm{2}: \\ $$$$\mathrm{tip}\:\mathrm{1}: \\ $$$${xy}−\mathrm{1}\:=\:\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+\left({x}−\mathrm{1}\right)+\left({y}−\mathrm{1}\right) \\ $$$$\mathrm{tip}\:\mathrm{2}: \\ $$$$\left(\mathrm{1}+{x}_{\mathrm{1}} \right)\left(\mathrm{1}+{x}_{\mathrm{2}} \right)\:=\:\mathrm{1}+\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{using}\:\mathrm{triangle}\:\mathrm{inequality}\:\mathrm{and}\:\mathrm{tip}\:\mathrm{1}: \\ $$$$\mid{xy}−\mathrm{1}\mid\:=\:\mid\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+\left({x}−\mathrm{1}\right)+\left({y}−\mathrm{1}\right)\mid \\ $$$$\leqslant\:\mid\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\mid+\mid\left({x}−\mathrm{1}\right)+\left({y}−\mathrm{1}\right)\mid \\ $$$$\leqslant\:\mid\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$=\:\mid\mathrm{1}−{x}\mid\mid\mathrm{1}−{y}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$=\:\mid{x}−\mathrm{1}\mid\mid{y}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$ \\ $$$$\rightarrow\:\mid{xy}−\mathrm{1}\mid\leqslant\mid{x}−\mathrm{1}\mid\mid{y}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$\overset{+\mathrm{1}} {\rightarrow}\:\mid{xy}−\mathrm{1}\mid+\mathrm{1}\:\leqslant\mid{x}−\mathrm{1}\mid\mid{y}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid+\mathrm{1} \\ $$$$\mathrm{using}\:\mathrm{tip}\:\mathrm{2}\:\mathrm{to}\:\mathrm{RHS}: \\ $$$$\mid{xy}−\mathrm{1}\mid+\mathrm{1}\leqslant\left(\mathrm{1}+\mid{x}−\mathrm{1}\mid\right)\left(\mathrm{1}+\mid{y}−\mathrm{1}\mid\right) \\ $$$$ \\ $$
Commented by mathocean1 last updated on 22/Jan/22
$${thanks}\:{sir}. \\ $$