Question Number 194448 by York12 last updated on 07/Jul/23 $${If}\:{a}\:,\:{b}\:,\:{c}\:>\mathrm{0}\:,\:{such}\:{that}\:{a}+{b}+{c}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}}+\frac{\mathrm{1}}{\mathrm{1}+{ac}}+\frac{\mathrm{1}}{\mathrm{1}+{bc}}\geqslant\frac{\mathrm{9}}{\mathrm{2}\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194421 by York12 last updated on 06/Jul/23 $$ \\ $$$${If}\:{a}\:,\:{b}\:,\:{c}\:>\mathrm{0}\:,\:{such}\:{that}\:{a}+{b}+{c}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}}+\frac{\mathrm{1}}{\mathrm{1}+{ac}}+\frac{\mathrm{1}}{\mathrm{1}+{bc}}\geqslant\frac{\mathrm{9}}{\mathrm{2}\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)} \\ $$ Commented by York12 last updated on 07/Jul/23…
Question Number 194343 by BaliramKumar last updated on 04/Jul/23 $$\lfloor\mathrm{9}.\overset{−} {\mathrm{9}}\rfloor\:=\:? \\ $$ Answered by MM42 last updated on 04/Jul/23 $$\mathrm{9} \\ $$ Answered by…
Question Number 194312 by Mingma last updated on 03/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 194297 by York12 last updated on 02/Jul/23 $${Let}\:{a}\:,\:{b}\:,\:{c}\:{be}\:\:{real}\:{positive}\:{numbers}\:\&\: \\ $$$${abc}=\mathrm{1}\: \\ $$$${prove}\:{that} \\ $$$$\frac{{ab}}{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}}+\frac{{bc}}{{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +{bc}}+\frac{{ac}}{{a}^{\mathrm{5}} +{c}^{\mathrm{5}} +{ac}}\leqslant\mathrm{1} \\ $$ Answered…
Question Number 194158 by Frix last updated on 28/Jun/23 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}: \\ $$$$\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1} \\ $$$$\mathrm{With}\:{s},\:{t},\:{u},\:{v}\:\in\mathbb{N}\:\mathrm{and}\:{s}<{t}<{u}<{v} \\ $$ Answered by AST last updated on 29/Jun/23 $${s}<{t}\Rightarrow\frac{\mathrm{1}}{{s}}>\frac{\mathrm{1}}{{t}}\Rightarrow\frac{\mathrm{4}}{{s}}>\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1}\Rightarrow{s}<\mathrm{4} \\…
Question Number 194144 by mr W last updated on 28/Jun/23 $${fill}\:{with}\:{different}\:{natural}\:{numbers}: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)} \\ $$ Answered by York12 last updated on 28/Jun/23 $${egyptian}\:{fractions} \\ $$$$\frac{\mathrm{1}}{{k}}=\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}…
Question Number 194015 by Rupesh123 last updated on 26/Jun/23 Answered by witcher3 last updated on 26/Jun/23 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\mathrm{f}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \\ $$$${f}_{{k}+\mathrm{1}} ={f}_{{k}} +{f}_{{k}−\mathrm{1}} ;\mathrm{f}_{\mathrm{1}} =\mathrm{f}_{\mathrm{2}}…
Question Number 194021 by mr W last updated on 26/Jun/23 $${find}\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}=? \\ $$ Commented by BaliramKumar last updated on 26/Jun/23 $$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \: \\ $$…
Question Number 193853 by York12 last updated on 21/Jun/23 $$ \\ $$$$\boldsymbol{{Let}}\:\boldsymbol{{n}}\:\&\:\boldsymbol{{k}}\:\boldsymbol{{be}}\:\boldsymbol{{positive}}\:\boldsymbol{{integers}}\:\boldsymbol{{and}}\:\boldsymbol{{let}} \\ $$$$\boldsymbol{{S}}\:\boldsymbol{{be}}\:\boldsymbol{{a}}\:\boldsymbol{{set}}\:\boldsymbol{{of}}\:\boldsymbol{{n}}\:\boldsymbol{{points}}\:\boldsymbol{{in}}\:\boldsymbol{{The}}\:\boldsymbol{{plane}}\:\boldsymbol{{such}}\:\boldsymbol{{that}}\:: \\ $$$$\boldsymbol{{For}}\:\boldsymbol{{any}}\:\boldsymbol{{point}}\:\boldsymbol{{P}}\:\boldsymbol{{of}}\:\boldsymbol{{S}}\:\boldsymbol{{there}}\:\boldsymbol{{are}}\:\boldsymbol{{at}}\:\boldsymbol{{least}}\:\boldsymbol{{K}}\:\boldsymbol{{points}}\:\boldsymbol{{of}}\:\boldsymbol{{S}}\:\boldsymbol{{Equidistant}}\:\boldsymbol{{from}}\:\boldsymbol{{p}} \\ $$$$\boldsymbol{{Prove}}\:\boldsymbol{{that}}\:\boldsymbol{{k}}<\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{2}\boldsymbol{{n}}} \\ $$ Terms of Service Privacy Policy…