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Prima, a lui Cezar:

\( 1. \) Se considera \( a,b,c \) trei numere reale strict pozitive. Sa se arate ca \( \sum {\frac{{b + c}}{a}} \geq 3+ \frac{(a^2+b^2+c^2)(ab+bc+ca)}{abc(a+b+c)}.

\)

A doua, a d-lui Mircea Lascu:

\( 2. \) Fie \( a,b,c \) numere reale strict pozitive. Demonstrati inegalitatea \( \sum {\frac{a^2}{3a^2+b^2+2ac} \leq \frac{1}{2} \)

1. Dupa realizarea calculelor se ajunge la a demonstra:
\( 2(\sum_{cyc}{} {a^2b^2}) \) \( + \sum_{cyc}{} {(a^3b+b^3a)} + 2abc (\sum_{cyc}{} {a}) \) \( \geq \) \( 4abc (\sum_{cyc}{} {a}) \) \( + \sum_{cyc}{} {(a^3b+a^3c)} \) \( \Leftrightarrow \)

\( \Leftrightarrow \) \( \sum_{cyc}{} {a^2b^2} \) \( \geq \) \( abc(\sum_{cyc}{} {a}) \) \( \Leftrightarrow \) \( \sum_{cyc}{} {{(ab-bc)}^2} \) \( \geq 0 \) adevarat

2. \( \sum_{cyc}{} {\frac {a^2}{3a^2+b^2+2ac}} \) \( \leq \) \( \sum_{cyc}{} {\frac {a^2}{2a^2+2ab+2ac}} \) \( = \) \( \frac {1}{2} (\sum_{cyc}{} {\frac {a}{a+b+c}}) \) \( = \) \( \frac {1}{2} \)