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Question Number 195790 by justenspi last updated on 10/Aug/23
a,b,c are positive real numbers and abc =1 prove that (a−1+(1/b))(b−1+(1/c))(c−1+(1/a))≤1
$${a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers}\:{and}\:{abc}\:=\mathrm{1} \\ $$$${prove}\:{that} \\ $$$$\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\leqslant\mathrm{1} \\ $$
Commented by justenspi last updated on 10/Aug/23
Case(I)(a−1+(1/b)),(b−1+(1/c)),(c−1+(1/a))>0 ⇒(((a−1+(1/b))(b−1+(1/c))(c−1+(1/a))))^(1/3) ≤((a+b+c+ab+bc+ac−3)/3)......(i) ⇒(((a−1+(1/b))(b−1+(1/c))(c−1+(1/a))))^(1/3) ≤((ab+bc+ac−a−b−c+3)/3)......(ii) ⇒2(((a−1+(1/b))(b−1+(1/c))(c−1+(1/a))))^(1/3) ≤((2(ab+bc+ac))/3) ⇒(((a−1+(1/b))(b−1+(1/c))(c−1+(1/a))))^(1/3) ≤((ab+bc+ac)/3) the equality holds when a+b+c=ab+bc+ac∧abc=1∧a,b,c>0 ⇒a+b+c=(1/a)+(1/b)+(1/c)⇒a=1,b=1,c=1 hence the maximum value of (((a−1+(1/b))(b−1+(1/c))(c−1+(1/a))))^(1/3) →((1+1+1)/3)=1 ⇒(a−1+(1/b))(b−1+(1/c))(c−1+(1/a)) max =1 Case(II) At most one the factors is negative hence we are done
$${Case}\left({I}\right)\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right),\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right),\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)>\mathrm{0} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)}\leqslant\frac{{a}+{b}+{c}+{ab}+{bc}+{ac}−\mathrm{3}}{\mathrm{3}}……\left({i}\right) \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)}\leqslant\frac{{ab}+{bc}+{ac}−{a}−{b}−{c}+\mathrm{3}}{\mathrm{3}}……\left({ii}\right) \\ $$$$\Rightarrow\mathrm{2}\sqrt[{\mathrm{3}}]{\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)}\leqslant\frac{\mathrm{2}\left({ab}+{bc}+{ac}\right)}{\mathrm{3}} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)}\leqslant\frac{{ab}+{bc}+{ac}}{\mathrm{3}} \\ $$$${the}\:{equality}\:{holds}\:{when} \\ $$$${a}+{b}+{c}={ab}+{bc}+{ac}\wedge{abc}=\mathrm{1}\wedge{a},{b},{c}>\mathrm{0} \\ $$$$\Rightarrow{a}+{b}+{c}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\Rightarrow{a}=\mathrm{1},{b}=\mathrm{1},{c}=\mathrm{1} \\ $$$${hence}\:{the}\:{maximum}\:{value}\:{of} \\ $$$$\sqrt[{\mathrm{3}}]{\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)}\rightarrow\frac{\mathrm{1}+\mathrm{1}+\mathrm{1}}{\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\:{max}\:=\mathrm{1} \\ $$$${Case}\left({II}\right)\:{At}\:{most}\:{one}\:{the}\:{factors}\:{is}\:{negative} \\ $$$${hence}\:{we}\:{are}\:{done} \\ $$
Commented by justenspi last updated on 10/Aug/23
is that right , please help
$${is}\:{that}\:{right}\:,\:{please}\:{help}\: \\ $$
Answered by York12 last updated on 10/Aug/23
set a=(x/y),b=(y/z),c=(z/x) ⇒(a−1+(1/b))(b−1+(1/c))(c−1+(1/a)) =(((x−y+z)(x+y−z)(−x+y+z))/(xyz)) (√((x−y+z)(x+y−z)))≤((x−y+z+x+y−z)/2)=x....(i) (√((x+y−z)(−x+y+z)))≤((x+y−z−x+y+z)/2)=y....(ii) (√((x−y+z)(−x+y+z)))≤((x−y+z−x+y+z)/2)=z....(iii) ⇒(i)×(ii)×(iii)⇒(x−y+z)(x+y−z)(−x+y+z)≤xyz ⇒(((x−y+z)(x+y−z)(−x+y+z))/(xyz))≤1 ⇒(a−1+(1/b))(b−1+(1/c))(c−1+(1/a))≤1 In case one of the factors is negative ⇒(a−1+(1/b))(b−1+(1/c))(c−1+(1/a))≤0<1
$$ \\ $$$${set}\:{a}=\frac{{x}}{{y}},{b}=\frac{{y}}{{z}},{c}=\frac{{z}}{{x}} \\ $$$$\Rightarrow\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right) \\ $$$$=\frac{\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)\left(−{x}+{y}+{z}\right)}{{xyz}} \\ $$$$\sqrt{\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)}\leqslant\frac{{x}−{y}+{z}+{x}+{y}−{z}}{\mathrm{2}}={x}….\left({i}\right) \\ $$$$\sqrt{\left({x}+{y}−{z}\right)\left(−{x}+{y}+{z}\right)}\leqslant\frac{{x}+{y}−{z}−{x}+{y}+{z}}{\mathrm{2}}={y}….\left({ii}\right) \\ $$$$\sqrt{\left({x}−{y}+{z}\right)\left(−{x}+{y}+{z}\right)}\leqslant\frac{{x}−{y}+{z}−{x}+{y}+{z}}{\mathrm{2}}={z}….\left({iii}\right) \\ $$$$\Rightarrow\left({i}\right)×\left({ii}\right)×\left({iii}\right)\Rightarrow\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)\left(−{x}+{y}+{z}\right)\leqslant{xyz} \\ $$$$\Rightarrow\frac{\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)\left(−{x}+{y}+{z}\right)}{{xyz}}\leqslant\mathrm{1} \\ $$$$\Rightarrow\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\leqslant\mathrm{1} \\ $$$${In}\:{case}\:\:{one}\:{of}\:{the}\:{factors}\:{is}\:{negative} \\ $$$$\Rightarrow\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\leqslant\mathrm{0}<\mathrm{1}\: \\ $$
Commented by justenspi last updated on 10/Aug/23
thanks but is my solution right
$${thanks}\:{but}\:{is}\:{my}\:{solution}\:{right} \\ $$

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