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Question Number 117122 by Ar Brandon last updated on 09/Oct/20
Given a,b,c ∈R^3 such that abc=1. Show that: (a−1+(1/b))(b−1+(1/c))(c−1+(1/a))≤1
$$\mathrm{Given}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\in\mathbb{R}^{\mathrm{3}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{abc}=\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\left(\mathrm{a}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{b}}\right)\left(\mathrm{b}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{c}}\right)\left(\mathrm{c}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{a}}\right)\leqslant\mathrm{1} \\ $$
Commented by MJS_new last updated on 09/Oct/20
I don′t think it′s true for all (a∣b∣c)∈R^3
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:\left({a}\mid{b}\mid{c}\right)\in\mathbb{R}^{\mathrm{3}} \\ $$
Commented by MJS_new last updated on 09/Oct/20
let b=−a∧c=(1/(ab))=−(1/a^2 ) ⇒ (((a^2 +a+1)(a^2 −a+1)(a^2 −a+1))/a^3 )≤1 (1) ((a^6 −a^5 −2a^3 −a−1)/a^3 )≤0 but the numerator has got 2 real zeros a_1 ≈−.560769∧a_2 =−(1/a_1 )≈1.78326 ⇒ (1) is true for a≤a_1 ∨0<a≤a_2 and wrong for a_1 <a<0∨a>a_2
$$\mathrm{let}\:{b}=−{a}\wedge{c}=\frac{\mathrm{1}}{{ab}}=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\frac{\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} −{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} −{a}+\mathrm{1}\right)}{{a}^{\mathrm{3}} }\leqslant\mathrm{1}\:\left(\mathrm{1}\right) \\ $$$$\frac{{a}^{\mathrm{6}} −{a}^{\mathrm{5}} −\mathrm{2}{a}^{\mathrm{3}} −{a}−\mathrm{1}}{{a}^{\mathrm{3}} }\leqslant\mathrm{0} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{numerator}\:\mathrm{has}\:\mathrm{got}\:\mathrm{2}\:\mathrm{real}\:\mathrm{zeros} \\ $$$${a}_{\mathrm{1}} \approx−.\mathrm{560769}\wedge{a}_{\mathrm{2}} =−\frac{\mathrm{1}}{{a}_{\mathrm{1}} }\approx\mathrm{1}.\mathrm{78326} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{a}\leqslant{a}_{\mathrm{1}} \vee\mathrm{0}<{a}\leqslant{a}_{\mathrm{2}} \:\mathrm{and} \\ $$$$\:\:\:\:\:\:\mathrm{wrong}\:\mathrm{for}\:{a}_{\mathrm{1}} <{a}<\mathrm{0}\vee{a}>{a}_{\mathrm{2}} \\ $$
Commented by 1549442205PVT last updated on 10/Oct/20
For a=b=−3,c=1/9 then (a−1+(1/b))(b−1+(1/c))(c−1+(1/a))>1 Since L.H.S=(−4−(1/3))(−4+9)((1/9)−1−(1/3)) =((−13)/3)×5×(((−11)/9))=((715)/(27))>1
$$\mathrm{For}\:\mathrm{a}=\mathrm{b}=−\mathrm{3},\mathrm{c}=\mathrm{1}/\mathrm{9}\:\mathrm{then} \\ $$$$\:\:\:\:\:\left(\mathrm{a}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{b}}\right)\left(\mathrm{b}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{c}}\right)\left(\mathrm{c}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{a}}\right)>\mathrm{1} \\ $$$$\mathrm{Since}\:\mathrm{L}.\mathrm{H}.\mathrm{S}=\left(−\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(−\mathrm{4}+\mathrm{9}\right)\left(\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{−\mathrm{13}}{\mathrm{3}}×\mathrm{5}×\left(\frac{−\mathrm{11}}{\mathrm{9}}\right)=\frac{\mathrm{715}}{\mathrm{27}}>\mathrm{1} \\ $$
Commented by Ar Brandon last updated on 13/Oct/20
OK Sir, thanks for your suggestions

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