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Let-a-b-c-be-positive-real-numbers-with-sum-3-Prove-that-1-a-1-b-1-c-3-2a-2-bc-3-2b-2-ac-3-2c-2-ab-

Question Number 149244 by EDWIN88 last updated on 04/Aug/21 $$\:\:{Let}\:{a},{b},{c}\:{be}\:{positive}\:{real}\:{numbers} \\ $$$${with}\:{sum}\:\mathrm{3}.\:{Prove}\:{that}\: \\ $$$$\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}{a}^{\mathrm{2}} +{bc}}+\frac{\mathrm{3}}{\mathrm{2}{b}^{\mathrm{2}} +{ac}}+\frac{\mathrm{3}}{\mathrm{2}{c}^{\mathrm{2}} +{ab}} \\ $$ Terms of Service Privacy Policy Contact:…

Question-83694

Question Number 83694 by naka3546 last updated on 05/Mar/20 Commented by naka3546 last updated on 05/Mar/20 $${DE}\:=\:{EF}\:=\:{FC}\:, \\ $$$${BG}\:=\:\mathrm{2}\:{CG}\:, \\ $$$${ABCD}\:\:{is}\:\:{a}\:\:{square}\:. \\ $$$$\frac{\left[\:{BMN}\:\right]}{\left[\:{ABCD}\:\right]}\:\:=\:\:? \\ $$$$\left(\:{without}\:\:{Trigonometry}\:\:{or}\:{Menelause}\:\right)…

Question-18146

Question Number 18146 by mondodotto@gmail.com last updated on 15/Jul/17 Commented by prakash jain last updated on 16/Jul/17 $$\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \mathrm{3}{x}+\mathrm{9}{x}^{\mathrm{3}} +{x}}{\left(\mathrm{9}{x}^{\mathrm{2}}…

Question-18123

Question Number 18123 by mondodotto@gmail.com last updated on 15/Jul/17 Commented by prakash jain last updated on 15/Jul/17 $$\mathrm{log}\:\left({x}−\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} +\mathrm{log}\:\mathrm{5}−\mathrm{log}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} = \\ $$$$\mathrm{log}\:\mathrm{5}=\mathrm{log}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{log}\:\left({x}−\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{5}=\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}}…

Question-83638

Question Number 83638 by bshahid010@gmail.com last updated on 04/Mar/20 Answered by TANMAY PANACEA last updated on 05/Mar/20 $${S}=\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+…+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:\left({i}\:{think}\right) \\ $$$${S}=\underset{{n}=\mathrm{0}}…