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Replied by ex-member gbaloglou:

How about the sharp inequalities \((1+x_1)^{1/x_1}(1+x_2)^{1/x_2}\leq 4\) (proven), \((1+x_1)^{1/x_1}(1+x_2)^{1/x_2}(1+x_3)^{1/x_3}\leq 8\), and \((1+x_1)^{1/x_1}(1+x_2)^{1/x_2}...(1+x_n)^{1/x_n}<e^{n-1}\) (for \(n\geq 4\))?

Posted by ex-member gbaloglou : Prove or disprove: if \(x_1x_2...x_n=1\) with \(x_i>0\) then \((1+x_1)^{1/x_1}(1+x_2)^{1/x_2}...(1+x_n)^{1/x_n}\leq 2^n\) References: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=511549 , http://www.mathematica.gr/forum/viewtopic.php?f=56&t=3...

A solution by aziiri. \[e-\sum_{k=0}^{n} \frac{1}{k!} = \sum_{k=n+1}^{\infty}\frac{1}{k!}=\frac{1}{n!}\left( \frac{1}{n+1} +\frac{1}{(n+1)(n+2)} +\dots\right) \leq \frac{1}{n!} \sum_{k=1}^{\infty} \frac{1}{(n+1)^k} = \frac{1}{n\cdot n!}\] Since \(\displaystyle \sum_{k=0}^n \frac{n!}{k!} = p_n\in \m...

Solution by Gigaster Hello Vaggelis! We will show that \( S_y\) is continuous and that \(\exists x_0\in X\) such that \(\lim_{n\to\infty} S_{y}^{n}(z)=x_0\) for some \(z\in X\). From the above we can easily deduce that \( S_y\) has a fixed point since : \(\lim_{n\to\infty} S_y(S_{y}^{n}(z))=\lim_{n...

A second solution given by Prof. S. Ntougias: It is known that \begin{align*} & \cos{x}+\sin{x}=\frac{3}{5}\left({\cos{x}+2\,\sin{x}}\right)+\frac{1}{5}\left({2\,\cos{x}-\sin{x}}\right)\quad{\text{ and}}\\ & 5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}=1+\left({\sin{x}-2\,\cos{x}}\right)^2=6-\le...

A solution by Foteini Kaldi: \[\dfrac{1}{2}\cdot \dfrac{2}{3}\cdot \ldots \cdot\dfrac{2n}{2n+1}=\dfrac{1}{2n+1}\] but \[\dfrac{1}{2} <\dfrac{2}{3}\quad \Rightarrow \quad \left(\dfrac{1}{2}\right)^2 <\dfrac{1}{2}\cdot \dfrac{2}{3}\] \[\dfrac{3}{4} <\dfrac{4}{5}\quad \Rightarrow \quad \left(\dfrac{3}{...

Find \[\displaystyle\frac{13}{8}+\mathop{\sum}\limits_{n=0}^{\infty}\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!\,4^{2n+3}}\,.\]