Question Number 33096 by 7991 last updated on 10/Apr/18
$${z}\:{and}\:{w}\:\in\:\mathbb{C} \\ $$$${proof}\:\mid\mid{z}\mid−\mid{w}\mid\mid\:\leqslant\:\mid{z}−{w}\mid\:{and}\:\mid{z}\mid−\mid{w}\mid\leqslant\:\mid{z}+{w}\mid \\ $$
Answered by MJS last updated on 10/Apr/18
$${z}={x}+{y}\mathrm{i} \\ $$$${w}={u}+{v}\mathrm{i} \\ $$$$ \\ $$$$\mathrm{1}. \\ $$$$\mid\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\mid\leqslant\sqrt{\left({x}−{u}\right)^{\mathrm{2}} +\left({y}−{v}\right)^{\mathrm{2}} } \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\right)^{\mathrm{2}} \leqslant\left({x}−{u}\right)^{\mathrm{2}} +\left({y}−{v}\right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }+\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)\leqslant{x}^{\mathrm{2}} −\mathrm{2}{xu}+{u}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{yv}+{v}^{\mathrm{2}} \\ $$$$−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\leqslant−\mathrm{2}\left({xu}+{yv}\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\geqslant{xu}+{yv} \\ $$$$\mathrm{if}\:\left({xu}+{yv}\right)\leqslant\mathrm{0}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{if}\:\left({xu}+{yv}\right)>\mathrm{0}: \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)\geqslant\left({xu}+{yv}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {u}^{\mathrm{2}} +{x}^{\mathrm{2}} {v}^{\mathrm{2}} +{y}^{\mathrm{2}} {u}^{\mathrm{2}} +{y}^{\mathrm{2}} {v}^{\mathrm{2}} \geqslant{x}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{2}{xuyv}+{y}^{\mathrm{2}} {v}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {v}^{\mathrm{2}} −\mathrm{2}{xuyv}+{y}^{\mathrm{2}} {u}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left({xv}−{yu}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{true} \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{if}\:\mid{z}\mid−\mid{w}\mid\leqslant\mathrm{0}\:\mathrm{it}'\mathrm{s}\:\mathrm{already}\:\mathrm{done} \\ $$$$\mathrm{if}\:\mid{z}\mid−\mid{w}\mid>\mathrm{0}\:\mathrm{go}\:\mathrm{on}\:\mathrm{like}\:\mathrm{above} \\ $$