Beweise es mit vollst. Induktion über n.
\(\prod \limits_{k=1}^{n}\left(1+x_{k}\right)>1+\sum \limits_{k=1}^{n} x_{k} \)
Anfang mit n=2 .
\(\prod \limits_{k=1}^{2}\left(1+x_{k}\right) = \left(1+x_{1}\right)\cdot\left(1+x_{2}\right)\)
\( = 1+ x_{1} + x_{2} + x_{1} \cdot x_{2} >1+x_{1}+x_{2} = 1+\sum \limits_{k=1}^{2} x_{k} \)
weil die x'e alle positiv sind.
Ind.schritt: Es gelte für n dann hat man
\(\prod \limits_{k=1}^{n+1}\left(1+x_{k}\right)= \left(1+x_{k+1}\right)\cdot\prod \limits_{k=1}^{n}\left(1+x_{k}\right) \)
\( = \prod \limits_{k=1}^{n}\left(1+x_{k}\right) + x_{k+1}\cdot \prod \limits_{k=1}^{n}\left(1+x_{k}\right) \)
\( > 1+\sum \limits_{k=1}^{n} x_{k} + x_{k+1}\cdot ( 1+\sum \limits_{k=1}^{n} x_{k} ) \)
\( = 1+\sum \limits_{k=1}^{n} x_{k} + x_{k+1} + x_{k+1}\cdot \sum \limits_{k=1}^{n} x_{k} \)
\( = 1+\sum \limits_{k=1}^{n+1} x_{k} + x_{k+1}\cdot \sum \limits_{k=1}^{n} x_{k} \)
Und weil \( x_{k+1}\cdot \sum \limits_{k=1}^{n} x_{k} \) jedenfalls positiv ist, gilt auch
\( > 1+\sum \limits_{k=1}^{n+1} x_{k} \) q.e.d.
