Chủ đề này có 1 trả lời

[Super Fields]

Với $x_i>0;y_i>0$ và $i=\overline{1,n}$, chứng minh rằng:

 

$\sqrt[n]{x_{1}x_{2}..x_{n}}+\sqrt[n]{y_{1}y_{2}..y_{n}}\leq \sqrt[n]{(x_1+y_1)(x_2+y_2)..(x_n+y_n)}$

[Messi10597]

$dpcm\Leftrightarrow \sqrt[n]{\frac{x_{1}x_{2}...x_{n}}{(x_{1}+y_{1})(x_{2}+y_{2})...(x_{n}+y_{n})}}+\sqrt[n]{\frac{y_{1}y_{2}...y_{n}}{(x_{1}+y_{1})(x_{2}+y_{2})...(x_{n}+y_{n})}}\leq 1$

Áp dụng AM-GM ta có:

$\sqrt[n]{\frac{x_{1}x_{2}...x_{n}}{(x_{1}+y_{1})(x_{2}+y_{2})...(x_{n}+y_{n})}}\leq \frac{1}{n}\left ( \frac{x_{1}}{x_{1}+y_{1}} +\frac{x_{2}}{x_{2}+y_{2}}+...+\frac{x_{n}}{x_{n}+y_{n}}\right )$

$\sqrt[n]{\frac{y_{1}y_{2}...y_{n}}{(x_{1}+y_{1})(x_{2}+y_{2})...(x_{n}+y_{n})}}\leq \frac{1}{n}\left ( \frac{y_{1}}{x_{1}+y_{1}} +\frac{y_{2}}{x_{2}+y_{2}}+...+\frac{y_{n}}{x_{n}+y_{n}}\right )$

Cộng 2 BĐT trên ta có đpcm