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Let P(x) be the real polynomial function, P(x) = ax^3 + bx^2 + cx + d. Prove that if |P(x)| \leq 1 for all x such that |x| \leq 1, then

|a| + |b| + |c| + |d| \leq 7.

Slični zadaci

Let a_{1}, a_{2}...a_{n} be non-negative reals, not all zero. Show that that
(a) The polynomial p(x) = x^{n} - a_{1}x^{n - 1} + ... - a_{n - 1}x - a_{n} has preceisely 1 positive real root R.
(b) let A = \sum_{i = 1}^n a_{i} and B = \sum_{i = 1}^n ia_{i}. Show that A^{A} \leq R^{B}.
Let x,y and z be positive real numbers such that xyz=1. Prove that


\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)}  \geq \frac{3}{4}.
A game is played by n girls (n \geq 2), everybody having a ball. Each of the \binom{n}{2} pairs of players, is an arbitrary order, exchange the balls they have at the moment. The game is called nice nice if at the end nobody has her own ball and it is called tiresome if at the end everybody has her initial ball. Determine the values of n for which there exists a nice game and those for which there exists a tiresome game.
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}.
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}.
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