Question Number 125960 by AST last updated on 26/Sep/22
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Answered by mahdipoor last updated on 16/Dec/20
$$\begin{cases}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mid{a}\mid^{\mathrm{2}} +\mid{b}\mid^{\mathrm{2}} }\\{\mathrm{2}{ab}\leqslant\mathrm{2}\mid{a}\mid.\mid{b}\mid}\end{cases} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\leqslant\mid{a}\mid^{\mathrm{2}} +\mid{b}\mid^{\mathrm{2}} +\mathrm{2}\mid{a}\mid.\mid{b}\mid \\ $$$$\Rightarrow\left({a}+{b}\right)^{\mathrm{2}} \leqslant\left(\mid{a}\mid+\mid{b}\mid\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mid{a}+{b}\mid\leqslant\mid{a}\mid+\mid{b}\mid \\ $$
Commented by mahdipoor last updated on 16/Dec/20
$$\:{x}^{\mathrm{2}} \leqslant{y}^{\mathrm{2}} \:\Rightarrow\mid{x}\mid\leqslant\mid{y}\mid\: \\ $$
Commented by mahdipoor last updated on 16/Dec/20
$${in}\:{my}\:{ans};{x}={a}+{b}\:\:\:{y}=\mid{a}\mid+\mid{b}\mid \\ $$$$\mid{x}\mid=\mid{a}+{b}\mid\:\:\:\mid{y}\mid=\mid\mid{a}\mid+\mid{b}\mid\mid=\mid{a}\mid+\mid{b}\mid \\ $$
Answered by arcana last updated on 19/Dec/20
$$\mathrm{using}\:−\mid{a}\mid\leqslant{a}\leqslant\mid{a}\mid\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\mid{b}\mid\leqslant{b}\leqslant\mid{b}\mid, \\ $$$$\mathrm{then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mid{a}\mid−\mid{b}\mid\leqslant{a}+{b}\leqslant\mid{a}\mid+\mid{b}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:−\left(\mid{a}\mid+\mid{b}\mid\right)\leqslant{a}+{b}\leqslant\mid{a}\mid+\mid{b}\mid \\ $$$$\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\mid{a}+{b}\mid\leqslant\mid{a}\mid+\mid{b}\mid \\ $$