\( \left(\frac{\sum \limits_{k=1}^{n+1} a_{k}}{(n+1) \cdot \bar{a}_{n}}\right)^{n+1} \)
\( = \left(\frac{a_{n+1}+\sum \limits_{k=1}^{n} a_{k}}{(n+1) \cdot \bar{a}_{n}}\right)^{n+1} \)
\( = \left(\frac{a_{n+1}}{(n+1) \cdot \bar{a}_{n}}+\frac{\sum \limits_{k=1}^{n} a_{k}}{(n+1) \cdot \bar{a}_{n}}\right)^{n+1} \)
\( = \left(\frac{a_{n+1}}{(n+1) \cdot \bar{a}_{n}}+\frac{n \cdot \bar{a}_{n}}{(n+1) \cdot \bar{a}_{n}}\right)^{n+1} \)
\( = \left(\frac{a_{n+1}}{(n+1) \cdot \bar{a}_{n}}+\frac{n}{n+1}\right)^{n+1} \)
\( = \left(\frac{a_{n+1}}{(n+1) \cdot \bar{a}_{n}}+1-\frac{1}{n+1}\right)^{n+1} \)
\( = \left(1+\frac{a_{n+1}}{(n+1) \cdot \bar{a}_{n}}-\frac{ \bar{a}_{n}}{(n+1)\cdot \bar{a}_{n}}\right)^{n+1} \)
\( = \left(1+\frac{a_{n+1}-\bar{a}_{n}}{(n+1) \cdot \bar{a}_{n}} \right)^{n+1} \)
Jetzt Bernoulli gibt
\( \ge 1+(n+1)\frac{a_{n+1}-\bar{a}_{n}}{(n+1) \cdot \bar{a}_{n}} \)
\( = 1+\frac{a_{n+1}-\bar{a}_{n}}{ \bar{a}_{n}} \)
\( = 1+\frac{a_{n+1}}{ \bar{a}_{n}}-\frac{\bar{a}_{n}}{ \bar{a}_{n}} \)
\( = \frac{a_{n+1}}{ \bar{a}_{n}} \)
