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Question Number 188646 by Rupesh123 last updated on 04/Mar/23

Answered by Frix last updated on 04/Mar/23

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

Commented by Rupesh123 last updated on 04/Mar/23

Nice, sir!

Answered by mehdee42 last updated on 04/Mar/23

$$step1$$$$\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}=2\Rightarrow$$$$\left(\frac{1}{\sqrt{ab}}\right)\left(\frac{1}{\sqrt{ac}}\right)\left(\frac{1}{\sqrt{bc}}\right)⩽{\left(\frac{2}{3}\right)}^3\Rightarrow abc⩾1$$$$step2$$$$(2\sqrt{ab}-1)(2\sqrt{ac}-1)(2\sqrt{bc}-1)=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})(\sqrt{b}+\sqrt{c})⩾$$$$⩾\left(2\sqrt{ab}\right)\left(2\sqrt{ac}\right)\left(2\sqrt{bc}\right)=8abc⩾8$$$$\Rightarrow (\sqrt{ab}-\frac{1}{2})(\sqrt{ac}-\frac{1}{2})(\sqrt{bc}-\frac{1}{2})⩾1$$$$$$$$$$

Commented by Rupesh123 last updated on 04/Mar/23

Nice, sir!

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