Školjka

%V0 Suppose that $x_1, x_2, x_3, \ldots$ are positive real numbers for which $$x^n_n = \sum^{n-1}_{j=0} x^j_n$$ for $n = 1, 2, 3, \ldots$ Prove that $\forall n,$ $$2 - \frac{1}{2^{n-1}} \leq x_n < 2 - \frac{1}{2^n}.$$

%V0 For $n \geq 3$ and $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$ given real numbers we have the following instructions: - place out the numbers in some order in a ring; - delete one of the numbers from the ring; - if just two numbers are remaining in the ring: let $S$ be the sum of these two numbers. Otherwise, if there are more the two numbers in the ring, replace Afterwards start again with the step (2). Show that the largest sum $S$ which can result in this way is given by the formula $$S_{max}= \sum^n_{k=2} \begin{pmatrix} n -2 \\ [\frac{k}{2}] - 1\end{pmatrix}a_{k}.$$

%V0 Consider two monotonically decreasing sequences $\left( a_k\right)$ and $\left( b_k\right)$, where $k \geq 1$, and $a_k$ and $b_k$ are positive real numbers for every k. Now, define the sequences $c_k = \min \left( a_k, b_k \right)$; $A_k = a_1 + a_2 + ... + a_k$; $B_k = b_1 + b_2 + ... + b_k$; $C_k = c_1 + c_2 + ... + c_k$ for all natural numbers k. (a) Do there exist two monotonically decreasing sequences $\left( a_k\right)$ and $\left( b_k\right)$ of positive real numbers such that the sequences $\left( A_k\right)$ and $\left( B_k\right)$ are not bounded, while the sequence $\left( C_k\right)$ is bounded? (b) Does the answer to problem (a) change if we stipulate that the sequence $\left( b_k\right)$ must be $\displaystyle b_k = \frac {1}{k}$ for all k ?

%V0 Let $n$ be a positive integer and let $(x_1,\ldots,x_n)$, $(y_1,\ldots,y_n)$ be two sequences of positive real numbers. Suppose $(z_2,\ldots,z_{2n})$ is a sequence of positive real numbers such that $z_{i+j}^2 \geq x_iy_j \qquad$ for all $1\le i,j \leq n$. Let $M=\max\{z_2,\ldots,z_{2n}\}$. Prove that $$\biggl(\frac{M+z_2+\cdots+z_{2n}}{2n}\biggr)^2\ge \biggl(\frac{x_1+\cdots+x_n}{n}\biggr)\biggl(\frac{y_1+\cdots+y_n}{n}\biggr...$$ comment Edited by Orl.

%V0 If $a, b, c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that $$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.$$

%V0 Let $n$ be a positive integer, and let $x$ and $y$ be a positive real number such that $x^n + y^n = 1.$ Prove that $$ \left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right) < \frac{1}{(1 - x)(1 - y)} \text{.}$$ Author: unknown author, Estonia