Cho các số thực không âm $a_{i}(i=\overline{1,n}), n\in N^{*})$, chứng minh rằng: $\frac{1}{n}\sum_{i=1}^{n}a_{i}^{m}\geq \left ( \frac{1}{n}\sum_{i=1}^{n}a_{i} \right )^m$
Phải thêm $m\in N^{*}$ chứ !!!
Xét $2$ trường hợp
$\bigstar$ Nếu $m$ chẵn ta có:
$\frac{1}{n}\sum_{i=1}^{n}a_{i}^{m}\geq \frac{1}{n^{2}}\left ( \sum_{i=1}^{n}\sqrt{a_{i}^{m}} \right )^{2}\geq \frac{1}{n^{m}}\left ( \sum_{i=1}^{n}a_{i} \right )^{m}=\left ( \frac{1}{n}\sum_{i=1}^{n}a_{i} \right )^{m}$
$\bigstar$ Nếu $m$ lẻ ta có:
$\frac{1}{n}\sum_{i=1}^{n}a_{i}^{m}=\sum_{i=1}^{n}\frac{a_{i}^{m+1}}{n.a_{i}}\geq \frac{1}{n.\sum_{i=1}^{n}a_{i}}\left ( \sum_{i=1}^{n}\sqrt{a_{i}^{m+1}} \right )^{2}\geq \frac{1}{n^{m}\sum_{i=1}^{n}a_{i}}\left ( \sum_{i=1}^{n}a_{i} \right )^{m+1}=\frac{1}{n^{m}}\left ( \sum_{i=1}^{n}a_{i} \right )^{m}=\left ( \frac{1}{n}\sum_{i=1}^{n}a_{i} \right )^{m}$