radio wrote: Let $ a,b,c > 0$. Prove that $ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\geq 3\sqrt {\frac {a^2 + b^2 + c^2}{ab + bc + ca}}$ We can suppose that $ a \geq b \geq c$ then we have $ \sum_{cyc} \frac {b}{a} \geq \sum_{cyc} \frac {a}{b}$ Note that we can prove: $ \sum_{cyc} \frac {a^2}{b^2} + 6 \geq \frac {3}{2}\sum_{cyc} \frac {a^2 + b^2}{ab}$ So $ (\sum_{cyc} \frac {a}{b})^2 = \sum_{cyc} \frac {a^2}{b^2} + 2\sum_{cyc} \frac {b}{a} \geq \frac {3}{2}\sum_{cyc} \frac {a^2 + b^2}{ab} - 6 + \sum_{cyc} \frac {a}{b} + \sum_{cyc} \frac {b}{a} = \frac {5}{2}\sum_{cyc} \frac {a^2 + b^2}{ab} - 6$ We need prove that: $ \frac {5}{2}\sum_{cyc} \frac {a^2 + b^2}{ab} - 6 \geq \frac {9(a^2 + b^2 + c^2)}{ab + bc + ca}$ or $ \sum_{cyc} (a - b)^2(\frac {5c}{a} + \frac {5c}{b} - 4) \geq 0$ Note $ a \geq b \geq c$ so $ \frac {5a}{b} + \frac {5a}{c} - 4 \geq \frac{5b}{c} + \frac {5b}{a} - 4 \geq \frac {5c}{a} + \frac {5c}{b} - 4$ Done!

chien than wrote: radio wrote: Let $ a,b,c > 0$. Prove that $ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\geq 3\sqrt {\frac {a^2 + b^2 + c^2}{ab + bc + ca}}$ We can suppose that $ a \geq b \geq c$ then we have $ \sum_{cyc} \frac {b}{a} \geq \sum_{cyc} \frac {a}{b}$ Note that we can prove: $ \sum_{cyc} \frac {a^2}{b^2} + 6 \geq \frac {3}{2}\sum_{cyc} \frac {a^2 + b^2}{ab}$ So $ (\sum_{cyc} \frac {a}{b})^2 = \sum_{cyc} \frac {a^2}{b^2} + 2\sum_{cyc} \frac {b}{a} \geq \frac {3}{2}\sum_{cyc} \frac {a^2 + b^2}{ab} - 6 + \sum_{cyc} \frac {a}{b} + \sum_{cyc} \frac {b}{a} = \frac {5}{2}\sum_{cyc} \frac {a^2 + b^2}{ab} - 6$ We need prove that: $ \frac {5}{2}\sum_{cyc} \frac {a^2 + b^2}{ab} - 6 \geq \frac {9(a^2 + b^2 + c^2)}{ab + bc + ca}$ or $ \sum_{cyc} (a - b)^2(\frac {5c}{a} + \frac {5c}{b} - 4) \geq 0$ Note $ a \geq b \geq c$ so $ \frac {5a}{b} + \frac {5a}{c} - 4 \geq \frac {5b}{c} + \frac {5b}{a} - 4 \geq \frac {5c}{a} + \frac {5c}{b} - 4$ Done! you can't suppose $ a \geq b \geq c$ because the initial inequality isn't symmetric..

radio wrote: Let $ a,b,c > 0$. Prove that $ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\geq 3\sqrt {\frac {a^2 + b^2 + c^2}{ab + bc + ca}}$ after clearing denominators,It's equivalent to prove that \[ \sum_{cyc}a^5bc^2+\sum_{cyc}a^5c^3+\sum_{cyc}a^4bc^3+2\sum_{cyc}a^4b^3c+2\sum_{cyc}a^3b^3c^2 \ge 7\sum_{cyc}a^4b^2c^2\] Which is obviously true by AM-GM;\[ a^5bc^2+a^5c^3+a^4bc^3+a^4b^3c+a^4b^3c+a^3b^3c^2+a^3b^3c^2 \ge 7a^4b^2c^2\]

radio wrote: Let $ a,b,c>0$. Prove that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ Mongolian 2007