khanhsy New member Xu 0 6/12/12 #2 tieukhanhlinh nói: Nhấn để mở rộng... Áp dụng \[Bunhiacopxki\] ta có. \[\sum_{cyclic}\sqrt{\(1^2+9^2\).\(x^2+\frac{1}{x^2}\)}\ge \sum_{cyclic}\(x+\frac{9}{x}\)\] \[\righ \sum_{cyclic}\sqrt{\(x^2+\frac{1}{x^2}\)}\ge \frac{1}{\sqrt{82}}\sum_{cyclic}\(x+\frac{9}{x} \)\] \[\righ \sum_{cyclic}\sqrt{\(x^2+\frac{1}{x^2}\)}\ge \frac{1}{\sqrt{82}}\sum_{cyclic}\(81x+\frac{9}{x}-80x \)(1)\] Ta lại có theo \[AM-GM\] \[81x+\frac{9}{x}\ge 54\] \[\righ\sum_{cyclic}\(81x+\frac{9}{x}\)\ge 162 (2)\] \[(1)&(2)\righ \sum_{cyclic}\sqrt{\(x^2+\frac{1}{x^2}\)}\ge \frac{1}{\sqrt{82}} .\(162-80.\sum_{cyclic}a\)\ge \sqrt{82}\ \(dpcm\) \]
tieukhanhlinh nói: Nhấn để mở rộng... Áp dụng \[Bunhiacopxki\] ta có. \[\sum_{cyclic}\sqrt{\(1^2+9^2\).\(x^2+\frac{1}{x^2}\)}\ge \sum_{cyclic}\(x+\frac{9}{x}\)\] \[\righ \sum_{cyclic}\sqrt{\(x^2+\frac{1}{x^2}\)}\ge \frac{1}{\sqrt{82}}\sum_{cyclic}\(x+\frac{9}{x} \)\] \[\righ \sum_{cyclic}\sqrt{\(x^2+\frac{1}{x^2}\)}\ge \frac{1}{\sqrt{82}}\sum_{cyclic}\(81x+\frac{9}{x}-80x \)(1)\] Ta lại có theo \[AM-GM\] \[81x+\frac{9}{x}\ge 54\] \[\righ\sum_{cyclic}\(81x+\frac{9}{x}\)\ge 162 (2)\] \[(1)&(2)\righ \sum_{cyclic}\sqrt{\(x^2+\frac{1}{x^2}\)}\ge \frac{1}{\sqrt{82}} .\(162-80.\sum_{cyclic}a\)\ge \sqrt{82}\ \(dpcm\) \]