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Category: Algebra

For-a-b-c-0-if-a-b-c-n-determine-minimum-and-maximum-values-of-a-2-b-2-c-2-ab-bc-ca-

Question Number 32767 by Rasheed.Sindhi last updated on 01/Apr/18 $$\mathrm{For}\:{a},{b},{c}\geqslant\mathrm{0}\:\mathrm{if}\:{a}+{b}+{c}={n},\:\mathrm{determine} \\ $$$$\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{values}\:\mathrm{of} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}. \\ $$ Commented by Rasheed.Sindhi last updated on…

Let-a-n-be-a-sequence-such-that-a-1-2-a-n-1-3a-n-4-2a-n-3-n-1-find-a-n-

Question Number 98280 by I want to learn more last updated on 12/Jun/20 $$\boldsymbol{\mathrm{Let}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{be}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\:\:\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:=\:\:\mathrm{2}, \\ $$$$\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}} \:\:=\:\:\frac{\mathrm{3}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:+\:\:\mathrm{4}}{\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:\:+\:\:\mathrm{3}},\:\:\:\:\:\boldsymbol{\mathrm{n}}\:\geqslant\:\mathrm{1},\:\:\:\:\:\boldsymbol{\mathrm{find}}\:\:\:\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \\ $$ Answered by…

let-p-x-be-a-polynomial-function-of-n-1-th-degree-and-p-k-k-for-k-1-2-3-n-find-p-0-and-p-n-1-example-n-10-

Question Number 98268 by mr W last updated on 16/Jun/20 $${let}\:{p}\left({x}\right)\:{be}\:{a}\:{polynomial}\:{function}\:{of} \\ $$$$\left({n}−\mathrm{1}\right)^{{th}} \:{degree}\:{and} \\ $$$${p}\left({k}\right)={k}\:{for}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3},…,{n} \\ $$$${find}\:{p}\left(\mathrm{0}\right)\:{and}\:{p}\left({n}+\mathrm{1}\right). \\ $$$${example}:\:{n}=\mathrm{10} \\ $$ Commented by MJS…

a-b-gt-0-a-2-b-2-1-prove-that-1-a-1-b-b-a-2-1-a-b-2-1-8-3-

Question Number 98267 by  M±th+et+s last updated on 12/Jun/20 $$\forall\:{a},{b}>\mathrm{0}\:,\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$${prove}\:{that} \\ $$$$\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left(\frac{{b}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{a}}{{b}^{\mathrm{2}} +\mathrm{1}}\right)\geqslant\frac{\mathrm{8}}{\mathrm{3}} \\ $$ Answered by maths mind last…