Question Number 188646 by Rupesh123 last updated on 04/Mar/23
Answered by Frix last updated on 04/Mar/23
![Due to symmetry the minimum of ((√(ab))−(1/2))((√(ac))−(1/2))((√(bc))−(1/2)) is at c=b=a ⇒ 3(√a)=2a(√a) ⇒ a=(3/2) The minimum is 1 [btw. (√u)=(((√v)+(√w))/(2(√(vw))−1)) ⇒ (√(ab)), (√(ac)), (√(bc))>(1/2)]](https://www.tinkutara.com/question/Q188649.png)
$$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\left(\sqrt{{ab}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{ac}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{bc}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{at} \\ $$$${c}={b}={a} \\ $$$$\Rightarrow \\ $$$$\mathrm{3}\sqrt{{a}}=\mathrm{2}{a}\sqrt{{a}}\:\Rightarrow\:{a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{1} \\ $$$$\left[\mathrm{btw}.\:\sqrt{{u}}=\frac{\sqrt{{v}}+\sqrt{{w}}}{\mathrm{2}\sqrt{{vw}}−\mathrm{1}}\:\Rightarrow\:\sqrt{{ab}},\:\sqrt{{ac}},\:\sqrt{{bc}}>\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$
Commented by Rupesh123 last updated on 04/Mar/23
Nice, sir!
Answered by mehdee42 last updated on 04/Mar/23
$${step}\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{ab}}}+\frac{\mathrm{1}}{\:\sqrt{{ac}}}+\frac{\mathrm{1}}{\:\sqrt{{bc}}}=\mathrm{2}\Rightarrow \\ $$$$\left(\frac{\mathrm{1}}{\:\sqrt{{ab}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{{ac}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{{bc}}}\right)\leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \Rightarrow{abc}\geqslant\mathrm{1} \\ $$$${step}\mathrm{2} \\ $$$$\left(\mathrm{2}\sqrt{{ab}}−\mathrm{1}\right)\left(\mathrm{2}\sqrt{{ac}}−\mathrm{1}\right)\left(\mathrm{2}\sqrt{{bc}}−\mathrm{1}\right)=\left(\sqrt{{a}}+\sqrt{{b}}\right)\left(\sqrt{{a}}+\sqrt{{c}}\right)\left(\sqrt{{b}}+\sqrt{{c}}\right)\geqslant \\ $$$$\geqslant\left(\mathrm{2}\sqrt{{ab}}\right)\left(\mathrm{2}\sqrt{{ac}}\right)\left(\mathrm{2}\sqrt{{bc}}\right)=\mathrm{8}{abc}\geqslant\mathrm{8} \\ $$$$\Rightarrow\left(\sqrt{{ab}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{ac}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{bc}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\geqslant\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 04/Mar/23
Nice, sir!