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Let a_0, a_1, a_2, \ldots be an arbitrary infinite sequence of positive numbers. Show that the inequality 1 + a_n > a_{n-1} \sqrt[n]{2} holds for infinitely many positive integers n.

Slični zadaci

Let a, b, c be positive real numbers such that \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c. Prove that:
\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}


Proposed by Juhan Aru, Estonia
(i) If x, y and z are three real numbers, all different from 1, such that xyz = 1, then prove that

\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1.
(With the \sum sign for cyclic summation, this inequality could be rewritten as \sum \frac {x^{2}}{\left(x - 1\right)^{2}} \geq 1.)

(ii) Prove that equality is achieved for infinitely many triples of rational numbers x, y and z.

Author: Walther Janous, Austria
Prove the inequality:

\displaystyle \sum_{i < j}{\frac {a_{i}a_{j}}{a_{i} + a_{j}}}\leq \frac {n}{2(a_{1} + a_{2} + ... + a_{n})}\cdot \sum_{i < j}{a_{i}a_{j}}

for positive reals a_{1}, a_{2}, ..., a_{n}.
Let x_1,x_2,\ldots,x_n be arbitrary real numbers. Prove the inequality

\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + \cdots + x_n^2} < \sqrt{n}.
Let a_{1},a_{2},\ldots ,a_{n} be positive real numbers such that a_{1}+a_{2}+\cdots +a_{n}<1. Prove that

\frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_1)(1 - a_2) \cdots (1 - a_n)} \leqslant \frac{1}{n^{n+1}}
Let n be an integer, n \geq 3. Let a_1, a_2, \ldots, a_n be real numbers such that 2 \leq a_i \leq 3 for i = 1, 2, \ldots, n. If s = a_1 + a_2 + \ldots + a_n, prove that \frac{a^{2}_{1}+a^{2}_{2}-a^{2}_{3}}{a_{1}+a_{2}-a_{3}}+\frac{a^{2}_{2}+a^{2}_{3}-a^{2}_{4}}{a_{2}+a_{3}-a_{4}}+\ldots+\frac{a^{2}_{n}+a^{2}_{1}-a^{2}_{2}}{a_{n}+a_{1}-a_{2}}\leq 2s-2n.