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Sa se arate ca pentru orice numere reale pozitive, \( a, b, c \) are loc inegalitatea

\( \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{a+b}{c}+\frac{c+a}{b}+\frac{b+c}{a}\right) \).

\( (\sum_{cyc}{} {\frac {a}{b}})^2 \geq \frac {3}{2} \sum_{cyc}{} ({\frac {a}{b} + \frac {a}{c}}) \) \( \Leftrightarrow \sum_{cyc}{} {\frac {a^2}{b^2}} + 2 \sum_{cyc}{} {\frac {a}{c}} \geq \frac {3}{2} (\sum_{cyc}{} {\frac {a}{b}} + \sum_{cyc}{} {\frac {a}{c}}) \) \( \Leftrightarrow 2 \sum_{cyc}{} {\frac {a^2}{b^2}} + \sum_{cyc}{} {\frac {a}{c}} \geq 3 \sum_{cyc}{} {\frac {a}{b}} \)

dar \( \frac {a^2}{b^2}+ \frac {b}{a} +\frac {b}{a} \geq 3 \) \( \Rightarrow \frac {1}{2} \sum_{cyc}{} {\frac {a^2}{b^2}} + \sum_{cyc}{} {\frac {a}{c}} = \sum_{cyc}{} {\frac {1}{2}(\frac {a^2}{b^2} + \frac {b}{a} + \frac {b}{a})} \geq \frac {9}{2} \)

\( \Rightarrow 2(\sum_{cyc}{} {\frac {a^2}{b^2}}) + \sum_{cyc}{} {\frac {a}{c}} \geq 3(\sum_{cyc}{} {\frac {a}{b}) \Leftrightarrow \frac {3}{2} (\sum_{cyc}{} {\frac {a^2}{b^2}}) + \frac {9}{2} \geq 3(\sum_{cyc}{} {\frac {a}{b}) \)

\( \Leftrightarrow \sum_{cyc}{} ({\frac {a^2}{b^2} + 1}) \geq 2(\sum_{cyc}{} {\frac {a}{b}}) \Leftrightarrow \sum_{cyc}{} {({\frac {a}{b} -1})^2} \geq 0 \)

Notam \( \frac{a}{b}=x ,\frac{b}{c}=y , \frac{c}{a}=z \) si inegalitatea devine \( (\sum x)^2 \geq \frac{3}{2}\cdot(\sum x +\sum \frac{1}{x}) \)

Efectuind calculele, tinind cont ca xyz=1 si \( \sum x^2 \geq \frac{1}{3}\cdot(\sum x)^2 \) e suficient sa aratam ca \( t^2\geq 3\cdot t \) unde \( t=x+y+z \).

Ultima relatie e evidenta conform inegalitatii mediilor.