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Question Number 218658 by Nicholas666 last updated on 14/Apr/25

Answered by Nicholas666 last updated on 14/Apr/25

Commented by A5T last updated on 14/Apr/25

$$\mathrm{x}⩽3\mathrm{and}\mathrm{y}⩾1\mathrm{does}\mathrm{not}\mathrm{imply}\mathrm{xy}⩽3$$$$\mathrm{In}\mathrm{fact},\mathrm{xy}\mathrm{is}\mathrm{unbounded}\mathrm{given}\mathrm{those}\mathrm{conditions}.$$$$$$$$\mathrm{So},\mathrm{if}\mathrm{we}\mathrm{take}\mathrm{x}=\sum _{\mathrm{cyc}}\mathrm{ab}⩽3\mathrm{and}\mathrm{y}=\sum _{\mathrm{cyc}}\left(\frac{1}{1+\mathrm{a}+\mathrm{b}}\right)⩾1,$$$$\mathrm{this}\mathrm{does}\mathrm{not}\mathrm{imply}\left(\sum _{\mathrm{cyc}}\mathrm{ab}\right)\left(\sum _{\mathrm{cyc}}\frac{1}{1+\mathrm{a}+\mathrm{b}}\right)⩾3$$

Answered by A5T last updated on 14/Apr/25

$$4-\mathrm{c}=1+\mathrm{a}+\mathrm{b};4-\mathrm{b}=1+\mathrm{a}+\mathrm{c};4-\mathrm{a}=1+\mathrm{b}+\mathrm{c}$$$$1+\mathrm{a}+\mathrm{b}⩾3\sqrt[3]{\mathrm{ab}}\Rightarrow \frac{\mathrm{ab}}{1+\mathrm{a}+\mathrm{b}}⩽\frac{\mathrm{ab}}{3\sqrt[3]{\mathrm{ab}}}=\frac{1}{3}{\left(\mathrm{ab}\right)}^{\frac{2}{3}}$$$$\Rightarrow \mathrm{\Sigma }\sqrt{\frac{\mathrm{ab}}{1+\mathrm{a}+\mathrm{b}}}⩽\frac{\sqrt{3}}{3}\mathrm{\Sigma }{\left(\mathrm{ab}\right)}^{\frac{1}{3}}$$$$\mathrm{But}{\left[\frac{\mathrm{\Sigma }{\left(\mathrm{ab}\right)}^{\frac{1}{3}}}{3}\right]}^3⩽\frac{\mathrm{\Sigma }\left(\mathrm{ab}\right)}{3}=\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{3}⩽\frac{{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^2}{9}=1$$$$\Rightarrow \mathrm{\Sigma }{\left(\mathrm{ab}\right)}^{\frac{1}{3}}⩽3$$$$\Rightarrow \mathrm{\Sigma }\sqrt{\frac{\mathrm{ab}}{1+\mathrm{a}+\mathrm{b}}}⩽\frac{\sqrt{3}}{3}\times 3=\sqrt{3}$$$$\mathrm{Equality}\mathrm{when}\mathrm{a}=\mathrm{b}=\mathrm{c}=1$$

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