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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)]

$$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\left(\sqrt{{ab}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{ac}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{bc}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{at} \\ $$$${c}={b}={a} \\ $$$$\Rightarrow \\ $$$$\mathrm{3}\sqrt{{a}}=\mathrm{2}{a}\sqrt{{a}}\:\Rightarrow\:{a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{1} \\ $$$$\left[\mathrm{btw}.\:\sqrt{{u}}=\frac{\sqrt{{v}}+\sqrt{{w}}}{\mathrm{2}\sqrt{{vw}}−\mathrm{1}}\:\Rightarrow\:\sqrt{{ab}},\:\sqrt{{ac}},\:\sqrt{{bc}}>\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$

Commented by Rupesh123 last updated on 04/Mar/23

Nice, sir!

Answered by mehdee42 last updated on 04/Mar/23

step1 (1/( (√(ab))))+(1/( (√(ac))))+(1/( (√(bc))))=2⇒ ((1/( (√(ab)))))((1/( (√(ac)))))((1/( (√(bc)))))≤((2/3))^3 ⇒abc≥1 step2 (2(√(ab))−1)(2(√(ac))−1)(2(√(bc))−1)=((√a)+(√b))((√a)+(√c))((√b)+(√c))≥ ≥(2(√(ab)))(2(√(ac)))(2(√(bc)))=8abc≥8 ⇒((√(ab))−(1/2))((√(ac))−(1/2))((√(bc))−(1/2))≥1

$${step}\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{ab}}}+\frac{\mathrm{1}}{\:\sqrt{{ac}}}+\frac{\mathrm{1}}{\:\sqrt{{bc}}}=\mathrm{2}\Rightarrow \\ $$$$\left(\frac{\mathrm{1}}{\:\sqrt{{ab}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{{ac}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{{bc}}}\right)\leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \Rightarrow{abc}\geqslant\mathrm{1} \\ $$$${step}\mathrm{2} \\ $$$$\left(\mathrm{2}\sqrt{{ab}}−\mathrm{1}\right)\left(\mathrm{2}\sqrt{{ac}}−\mathrm{1}\right)\left(\mathrm{2}\sqrt{{bc}}−\mathrm{1}\right)=\left(\sqrt{{a}}+\sqrt{{b}}\right)\left(\sqrt{{a}}+\sqrt{{c}}\right)\left(\sqrt{{b}}+\sqrt{{c}}\right)\geqslant \\ $$$$\geqslant\left(\mathrm{2}\sqrt{{ab}}\right)\left(\mathrm{2}\sqrt{{ac}}\right)\left(\mathrm{2}\sqrt{{bc}}\right)=\mathrm{8}{abc}\geqslant\mathrm{8} \\ $$$$\Rightarrow\left(\sqrt{{ab}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{ac}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\sqrt{{bc}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\geqslant\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Commented by Rupesh123 last updated on 04/Mar/23

Nice, sir!

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