Fie a ,b,c numere pozitive astfel incat abc=1. Demonstrati ca:
\( \sum {\frac{1}{a^2+2b^2+3}}\leq \frac{1}{2} \)
Mircea Lascu SHL 2006
Fie a,b,c numere...
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Marius Mainea
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Solutie
Avem inegalitatea :
\( a^2 + 2b^2 +3 \geq 2ab +b^2 + 3 = \frac{2}{c} + b^2 + 3 \) si analoagele acesteia .
Atunci :
\( \sum {\frac{1}{a^2+2b^2+3}}\leq \frac{c}{b^2c+3c+2} + \frac{a}{c^2a +3a +2} + \frac{b}{a^2b + 3b +2} \)
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Dar cum \( b^2c +3c +2 \geq 2bc + 2c +2 \) , atunci ramane de demonstrat ca :
\( \frac{c}{2bc + 2c + 2} + \frac{a}{2ca + 2a + 2} + \frac{b}{2ab + 2b + 2} \leq \frac{1}{2} \) , sau
\( \frac{c}{bc + c +1 } + \frac{a}{ca + a + 1} + \frac{b}{ab + b + 1} \leq 1 \) , pe care daca o prelucram , ajungem la :
\( \frac{1}{ab + b +1} + \frac{ab}{ab + b + 1} + \frac{b}{ab + b +1} = 1 \) .
Si inegalitatea este demonstrata .
\( a^2 + 2b^2 +3 \geq 2ab +b^2 + 3 = \frac{2}{c} + b^2 + 3 \) si analoagele acesteia .
Atunci :
\( \sum {\frac{1}{a^2+2b^2+3}}\leq \frac{c}{b^2c+3c+2} + \frac{a}{c^2a +3a +2} + \frac{b}{a^2b + 3b +2} \)
.
Dar cum \( b^2c +3c +2 \geq 2bc + 2c +2 \) , atunci ramane de demonstrat ca :
\( \frac{c}{2bc + 2c + 2} + \frac{a}{2ca + 2a + 2} + \frac{b}{2ab + 2b + 2} \leq \frac{1}{2} \) , sau
\( \frac{c}{bc + c +1 } + \frac{a}{ca + a + 1} + \frac{b}{ab + b + 1} \leq 1 \) , pe care daca o prelucram , ajungem la :
\( \frac{1}{ab + b +1} + \frac{ab}{ab + b + 1} + \frac{b}{ab + b +1} = 1 \) .
Si inegalitatea este demonstrata .