Question Number 205767 by lmcp1203 last updated on 30/Mar/24 $$ \\ $$$${a},{b},{c}\:\in\Re^{+} \:\: \\ $$$${a}+{b}+{c}=\mathrm{1} \\ $$$$\:\:\:{a}^{\mathrm{2}} /\left(\mathrm{1}+{b}+{c}\right)\:+\:{b}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{c}\right)\:\:+\:{c}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{b}\right)\geqslant{k} \\ $$$${find}\:\:\:{k}\:{max}. \\ $$$${hint}\::\:{inequality}\:{cauchy}\:{schwarz} \\…
Question Number 205770 by hardmath last updated on 30/Mar/24 $$\mathrm{If}\:\:\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{xyz}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\left(\sqrt{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{yz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{yz}\right)}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 205746 by Satyam1234 last updated on 29/Mar/24 Answered by A5T last updated on 29/Mar/24 $${C}\neq{J}+\mathrm{2}\Rightarrow{C}=\mathrm{3}{F} \\ $$$${A}\geqslant\mathrm{1};{A}=\mathrm{1}\Rightarrow{B}=\mathrm{3}\Rightarrow{I}=\mathrm{7}\Rightarrow{H}=\mathrm{4}\Rightarrow{A}=\mathrm{8}\:{X} \\ $$$${A}=\mathrm{2}\Rightarrow{B}=\mathrm{6}\Rightarrow{E}=\mathrm{2}\:{X} \\ $$$${A}=\mathrm{3}\Rightarrow{B}=\mathrm{9}\:{X}\left(\mathrm{6}\Rightarrow{B}<\mathrm{5}\:{or}\:{A}\geqslant\mathrm{4}\right)\:{or}\:{G}=\mathrm{0} \\ $$$$\:\left({G}=\mathrm{0}\Rightarrow{I}=\mathrm{3}\Rightarrow{D}=\mathrm{6}\Rightarrow{E}=\mathrm{18}\right){X}…
Question Number 205726 by hardmath last updated on 28/Mar/24 $$ \\ $$101 is chosen arbitrarily from the numbers 1,2,3,…,199,200. Prove that two of these selected…
Question Number 205733 by hardmath last updated on 28/Mar/24 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{6} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4a}^{\mathrm{2}} −\mathrm{9a}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4b}^{\mathrm{2}} −\mathrm{9b}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4c}^{\mathrm{2}} −\mathrm{9c}\:+\:\mathrm{6}}\:\leqslant\:\mathrm{0} \\ $$ Answered by…
Question Number 205680 by mnjuly1970 last updated on 27/Mar/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\lfloor{x}\:\rfloor\:+\:\lfloor\:{x}^{\mathrm{2}} \rfloor\:=\:\lfloor\:{x}^{\mathrm{3}} \:\rfloor \\ $$$$ \\ $$ Answered by Frix…
Question Number 205645 by hardmath last updated on 26/Mar/24 $$\mathrm{Find}:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{2}\:\mathrm{cosx}\:+\:\mathrm{3}\:\mathrm{sinx}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Frix last updated on 26/Mar/24 $${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{leads}\:\mathrm{to} \\ $$$$−\mathrm{4}\int\frac{{t}^{\mathrm{2}}…
Question Number 205672 by hardmath last updated on 26/Mar/24 Answered by Berbere last updated on 27/Mar/24 $$\Omega=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\sqrt{{k}\left({k}−\mathrm{1}\right)}}{\:\sqrt{\left(\sqrt{{k}−\mathrm{1}}+\sqrt{{k}}\right)^{\mathrm{2}} }}.\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{{k}}\right)^{−\mathrm{1}} \\ $$$${S}_{\mathrm{1}}…
Question Number 205643 by hardmath last updated on 26/Mar/24 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:=\:\mathrm{abc} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{bc}}\:+\:\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{ac}}\:+\:\frac{\mathrm{c}}{\mathrm{c}^{\mathrm{2}} \:+\:\mathrm{ab}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by A5T…
Question Number 205640 by hardmath last updated on 26/Mar/24 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{abc}\geqslant\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\geqslant\:\frac{\mathrm{1}+\mathrm{a}}{\mathrm{1}+\mathrm{b}}\:+\:\frac{\mathrm{1}+\mathrm{b}}{\mathrm{1}+\mathrm{c}}\:+\:\frac{\mathrm{1}+\mathrm{c}}{\mathrm{1}+\mathrm{a}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com