Question Number 196440 by mokys last updated on 24/Aug/23
$${prove}\:\mid{v}−{w}\mid\geqslant\mid{v}\mid−\mid{w}\mid \\ $$
Answered by Frix last updated on 24/Aug/23
$${v},\:{w}\:\in\mathbb{C} \\ $$$${v}={a}\mathrm{e}^{\mathrm{i}\alpha} \wedge{w}={b}\mathrm{e}^{\mathrm{i}\beta} \wedge{a},\:{b}\:\geqslant\mathrm{0} \\ $$$$\mid{v}−{w}\mid\geqslant\mid{v}\mid−\mid{w}\mid \\ $$$$\sqrt{{a}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\left(\alpha−\beta\right)\:+{b}^{\mathrm{2}} }\geqslant{a}−{b} \\ $$$$ \\ $$$$\mathrm{Case}\:\mathrm{1}:\:{a}−{b}<\mathrm{0} \\ $$$$\mathrm{True}\:\mathrm{because}\:\sqrt{…}\geqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{C}.\mathrm{ase}\:\mathrm{2}:\:{a}−{b}\geqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\left(\alpha−\beta\right)\:+{b}^{\mathrm{2}} \geqslant{a}^{\mathrm{2}} −\mathrm{2}{ab}−{b}^{\mathrm{2}} \\ $$$$−\mathrm{2}{ab}\mathrm{cos}\:\left(\alpha−\beta\right)\:\geqslant−\mathrm{2}{ab} \\ $$$$\mathrm{cos}\:\left(\alpha−\beta\right)\:\leqslant\mathrm{1}\:\mathrm{true} \\ $$
Answered by AST last updated on 25/Aug/23
$$\Leftrightarrow\left({v}−{w}\right)^{\mathrm{2}} \geqslant{v}^{\mathrm{2}} +{w}^{\mathrm{2}} −\mathrm{2}\mid{v}\mid\mid{w}\mid\Leftrightarrow{vw}\leqslant\mid{v}\mid\mid{w}\mid\left({true}\right) \\ $$