[Hỏi đáp] Chứng minh bất đắng thức

[tieukhanhlinh]

 

[khanhsy]

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}\ \(dpcm\) \]