Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 117122 by Ar Brandon last updated on 09/Oct/20

$$\mathrm{Given}\mathrm{a},\mathrm{b},\mathrm{c}\in {\mathbb{R}}^3\mathrm{such}\mathrm{that}\mathrm{abc}=1.\mathrm{Show}\mathrm{that}:$$$$(\mathrm{a}-1+\frac{1}{\mathrm{b}})(\mathrm{b}-1+\frac{1}{\mathrm{c}})(\mathrm{c}-1+\frac{1}{\mathrm{a}})⩽1$$

Commented by MJS_new last updated on 09/Oct/20

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{true}\mathrm{for}\mathrm{all}(a\mid b\mid c)\in {\mathbb{R}}^3$$

Commented by MJS_new last updated on 09/Oct/20

$$\mathrm{let}b=-a\wedge c=\frac{1}{ab}=-\frac{1}{a^2}$$$$\Rightarrow$$$$\frac{(a^2+a+1)(a^2-a+1)(a^2-a+1)}{a^3}⩽1\left(1\right)$$$$\frac{a^6-a^5-2a^3-a-1}{a^3}⩽0$$$$\mathrm{but}\mathrm{the}\mathrm{numerator}\mathrm{has}\mathrm{got}2\mathrm{real}\mathrm{zeros}$$$$a_1\approx -.560769\wedge a_2=-\frac{1}{a_1}\approx 1.78326$$$$\Rightarrow \left(1\right)\mathrm{is}\mathrm{true}\mathrm{for}a⩽a_1\vee 0<a⩽a_2\mathrm{and}$$$$\mathrm{wrong}\mathrm{for}{a}_1<a<0\vee a>a_2$$

Commented by 1549442205PVT last updated on 10/Oct/20

$$\mathrm{For}\mathrm{a}=\mathrm{b}=-3,\mathrm{c}=1/9\mathrm{then}$$$$(\mathrm{a}-1+\frac{1}{\mathrm{b}})(\mathrm{b}-1+\frac{1}{\mathrm{c}})(\mathrm{c}-1+\frac{1}{\mathrm{a}})>1$$$$\mathrm{Since}\mathrm{L}.\mathrm{H}.\mathrm{S}=(-4-\frac{1}{3})(-4+9)(\frac{1}{9}-1-\frac{1}{3})$$$$=\frac{-13}{3}\times 5\times \left(\frac{-11}{9}\right)=\frac{715}{27}>1$$

Commented by Ar Brandon last updated on 13/Oct/20

OK Sir, thanks for your suggestions

Terms of Service

Privacy Policy

Contact: info@tinkutara.com