Suma variabilelor este egala cu suma inverselor lor
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Suma variabilelor este egala cu suma inverselor lor
Fie \( a,b,c \) trei numere pozitive astfel incat \( a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \) . Daca \( a\le b\le c \), atunci \( ab^2c^3\ge 1 \).
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Marius Mainea
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Folosind ipoteza \( 3c\ge a+b+c=\frac{1}{a} +\frac{1}{b}+\frac{1}{c}\ge\frac{3}{c} \) deci \( c\ge 1 \)
Deasemenea \( 3a\le a+b+c=\frac{1}{a} +\frac{1}{b}+\frac{1}{c}\le\frac{3}{a} \) deci \( a\le 1 \)
Din relatia data \( a-\frac{1}{a}=(b+c)(\frac{1}{bc}-1) \) deci \( bc\ge1 \) (*)
\( c-\frac{1}{c}=(a+b)(\frac{1}{ab}-1) \) deci \( ab\le 1 \)
Deasamenea \( c-\frac{1}{c}=(a+b)(\frac{1}{ab}-1)\ge2\sqrt{ab}(\frac{1}{ab}-1)=2(\frac{1}{\sqrt{ab}}-\sqrt{ab})\ge\frac{1}{\sqrt{ab}}-\sqrt{ab} \) de unde \( c\ge\frac{1}{ \sqrt{ab}} \) (**)
Din (*) si (**) rezulta concluzia.
Deasemenea \( 3a\le a+b+c=\frac{1}{a} +\frac{1}{b}+\frac{1}{c}\le\frac{3}{a} \) deci \( a\le 1 \)
Din relatia data \( a-\frac{1}{a}=(b+c)(\frac{1}{bc}-1) \) deci \( bc\ge1 \) (*)
\( c-\frac{1}{c}=(a+b)(\frac{1}{ab}-1) \) deci \( ab\le 1 \)
Deasamenea \( c-\frac{1}{c}=(a+b)(\frac{1}{ab}-1)\ge2\sqrt{ab}(\frac{1}{ab}-1)=2(\frac{1}{\sqrt{ab}}-\sqrt{ab})\ge\frac{1}{\sqrt{ab}}-\sqrt{ab} \) de unde \( c\ge\frac{1}{ \sqrt{ab}} \) (**)
Din (*) si (**) rezulta concluzia.