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Prove-that-for-every-positive-real-numbers-x-y-z-and-xyz-1-hold-x-y-z-2-1-x-2-1-y-2-1-z-2-9-2-x-3-y-3-z-3-4-1-x-3-1-y-3-1-z-3-

Question Number 34031 by naka3546 last updated on 29/Apr/18 $${Prove}\:\:{that}\:\:{for}\:\:{every}\:\:{positive}\:\:{real}\:\:{numbers}\:\:{x},\:{y},\:{z}\:\:{and}\:\:\:{xyz}\:\:=\:\:\mathrm{1},\:\:{hold} \\ $$$$\left({x}\:+\:{y}\:+\:{z}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)\:\:\geqslant\:\:\mathrm{9}\:+\:\mathrm{2}\left({x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:+\:{z}^{\mathrm{3}} \right)\:+\:\mathrm{4}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)\: \\ $$ Terms…

Given-a-b-gt-0-and-n-N-u-n-v-n-gt-0-u-0-a-u-n-1-u-n-v-n-and-v-0-b-v-n-1-1-2-u-n-v-n-Show-that-the-sequences-u-n-and-u-n-are-convergent-and

Question Number 165081 by mathocean1 last updated on 25/Jan/22 $${Given}\:{a};\:{b}\:>\mathrm{0}\:{and}\:\forall\:{n}\:\in\:\mathbb{N},\:{u}_{{n}} ;{v}_{{n}} >\mathrm{0}\:. \\ $$$$\:\begin{cases}{{u}_{\mathrm{0}} ={a}}\\{{u}_{{n}+\mathrm{1}} =\sqrt{{u}_{{n}} {v}_{{n}} }}\end{cases}\:{and}\:\begin{cases}{{v}_{\mathrm{0}} ={b}}\\{{v}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{n}} +{v}_{{n}} \right).}\end{cases} \\ $$$${Show}\:{that}\:{the}\:{sequences}\:{u}_{{n}} \:{and}\:{u}_{{n}}…

4k-1-k-3-4k-1-k-3-11-k-3-1or11-I-can-t-understand-Who-can-help-me-

Question Number 34005 by math2018 last updated on 29/Apr/18 $$\frac{\mathrm{4}{k}+\mathrm{1}}{{k}+\mathrm{3}},\left(\mathrm{4}{k}+\mathrm{1},{k}+\mathrm{3}\right)=\left(\mathrm{11},{k}+\mathrm{3}\right)=\mathrm{1}{or}\mathrm{11}; \\ $$$${I}\:{can}'{t}\:{understand}.{Who}\:{can}\:{help}\:{me}? \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18 $${pls}\:{clarify}\:{the}\:{question}\:{or}\:{upload}\:{the}\:{image}\:{of}\:{question} \\ $$ Commented…

Question-99529

Question Number 99529 by Dwaipayan Shikari last updated on 21/Jun/20 Answered by mr W last updated on 21/Jun/20 $${a}={acceleration}\:{of}\:{m} \\ $$$${A}={acceleration}\:{of}\:{M} \\ $$$${ma}=−{mg}\frac{\mu}{\mathrm{2}} \\ $$$$\Rightarrow{a}=−\frac{\mu{g}}{\mathrm{2}}…

x-y-z-R-x-2-y-3-z-4-x-4-y-5-z-6-Prove-that-x-2-y-4-1-y-2-z-4-1-z-2-x-4-1-x-2-y-2-z-2-2-

Question Number 99513 by naka3546 last updated on 21/Jun/20 $${x},{y},{z}\:\:\in\:\:\mathbb{R}^{+} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} \:+\:{z}^{\mathrm{4}} \:\:=\:\:{x}^{\mathrm{4}} \:+\:{y}^{\mathrm{5}} \:+\:{z}^{\mathrm{6}} \\ $$$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{4}} +\mathrm{1}}\:\:+\:\:\frac{{y}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}\:\:+\:\:\frac{{z}^{\mathrm{2}}…