Question Number 78351 by loveineq. last updated on 16/Jan/20
$$\mathrm{Let}\:\:{a},{b},{c}\:\in\:\mathrm{R}^{+} \:\:\mathrm{and}\:\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{1}\:,\:\mathrm{where}\:\:\mathrm{0}\:<{b}\leqslant\:\mathrm{1}\:. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\frac{\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid}{\mathrm{2}}\:. \\ $$
Commented by loveineq. last updated on 19/Jan/20
$$\mathrm{More}\:\mathrm{stronger}\:\mathrm{is}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid\:. \\ $$
Answered by mind is power last updated on 16/Jan/20
$$\mathrm{x}−\mathrm{y}=\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right) \\ $$$$\Rightarrow\mid\mathrm{a}−\mathrm{b}\mid\mid\mathrm{b}−\mathrm{c}\mid=\mid\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}}\mid\mid\sqrt{\mathrm{b}}−\sqrt{\mathrm{c}}\mid\left(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}\right)\left(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}\right) \\ $$$$\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\:\:\geqslant\frac{\sqrt{\mathrm{x}+\mathrm{y}}}{\:\sqrt{\mathrm{2}}}\:\:\mathrm{proof}\Leftrightarrow\mathrm{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}\right)\geqslant\mathrm{x}+\mathrm{y} \\ $$$$\Leftrightarrow\mathrm{x}+\mathrm{y}+\mathrm{4}\sqrt{\mathrm{xy}}\geqslant\mathrm{0}\:\mathrm{True} \\ $$$$\left(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}\right)\left(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}\right)\geqslant\frac{\sqrt{\mathrm{a}+\mathrm{b}}.\sqrt{\mathrm{b}+\mathrm{c}}}{\mathrm{2}}=\frac{\sqrt{\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{b}+\mathrm{c}\right)}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mid\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}}\mid\mid\sqrt{\mathrm{b}}−\sqrt{\mathrm{c}}\mid\left(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}\right)\left(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}\right)\geqslant\frac{\mid\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}\mid}\mid\sqrt{\mathrm{b}}−\sqrt{\mathrm{c}}\mid}{\mathrm{2}} \\ $$
Answered by loveineq. last updated on 16/Jan/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by mind is power last updated on 19/Jan/20
$$\mathrm{y}'\mathrm{re}\:\mathrm{Welcom} \\ $$