Sa se arate ca pentru orice \( n\in\mathbb{N}^{*} \) si \( a_{1}, a_{2}, \ldots, a_{n}>0 \) are loc inegalitatea
\( a_{1}+\sqrt{a_{1}a_{2}}+\sqrt[3]{a_{1}a_{2}a_{3}}+\ldots +\sqrt[n]{a_{1}a_{2}\ldots a_{n}}\leq e(a_{1}+a_{2}+\ldots +a_{n}) \).
Inegalitatea 6, Carleman
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Inegalitatea 6, Carleman
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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pohoatza
Tare. Uite cum m-am gandit:
Cum \( \frac{(k+1)^{k}}{k!} \leq e^{k} \), avem imediat ca: \( \frac{1}{e} \cdot \sum_{k=1}^{n}{\left(\prod_{i=1}^{k}{a_{i}}\right)}^{\frac{1}{k}}\leq \sum_{k=1}^{n}{\left(\frac{(k+1)^{k}}{k!}\right)^{\frac{1}{k}}\left(\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}} \).
Dar \( \sum_{k=1}^{n}{\left(\frac{(k+1)^{k}}{k!}\right)^{\frac{1}{k}}\left(\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}}=\sum_{k=1}^{n}{\frac{1}{k+1}\left(k!\cdot\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}} \) , iar din GM-AM,
\( \sum_{k=1}^{n}{\frac{1}{k+1}\left(k!\cdot\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}} \leq \sum_{k=1}^{n}{\frac{a_{1}+2a_{2}+\ldots+ka_{k}}{k(k+1)}= \sum_{k=1}^{n}{\left(\frac{1}{k(k+1)} \sum_{i=1}^{k}{ia_{i}}\right)}=\sum_{i=1}^{n}{a_{i}} \).
Astfel, am obtinut deci ca \( \sum_{i=1}^{n}{a_{i}} \geq \frac{1}{e} \cdot \sum_{k=1}^{n}{\left(\prod_{i=1}^{k}{a_{i}}\right)}^{\frac{1}{k}} \), care este chiar inegalitatea initiala.
Cum \( \frac{(k+1)^{k}}{k!} \leq e^{k} \), avem imediat ca: \( \frac{1}{e} \cdot \sum_{k=1}^{n}{\left(\prod_{i=1}^{k}{a_{i}}\right)}^{\frac{1}{k}}\leq \sum_{k=1}^{n}{\left(\frac{(k+1)^{k}}{k!}\right)^{\frac{1}{k}}\left(\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}} \).
Dar \( \sum_{k=1}^{n}{\left(\frac{(k+1)^{k}}{k!}\right)^{\frac{1}{k}}\left(\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}}=\sum_{k=1}^{n}{\frac{1}{k+1}\left(k!\cdot\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}} \) , iar din GM-AM,
\( \sum_{k=1}^{n}{\frac{1}{k+1}\left(k!\cdot\prod_{i=1}^{k}{a_{i}}\right)^{\frac{1}{k}}} \leq \sum_{k=1}^{n}{\frac{a_{1}+2a_{2}+\ldots+ka_{k}}{k(k+1)}= \sum_{k=1}^{n}{\left(\frac{1}{k(k+1)} \sum_{i=1}^{k}{ia_{i}}\right)}=\sum_{i=1}^{n}{a_{i}} \).
Astfel, am obtinut deci ca \( \sum_{i=1}^{n}{a_{i}} \geq \frac{1}{e} \cdot \sum_{k=1}^{n}{\left(\prod_{i=1}^{k}{a_{i}}\right)}^{\frac{1}{k}} \), care este chiar inegalitatea initiala.