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Find all functions f: \mathbb{R}^{ + }\to\mathbb{R}^{ + } satisfying f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right) for all pairs of positive reals x and y. Here, \mathbb{R}^{ + } denotes the set of all positive reals.

Proposed by Paisan Nakmahachalasint, Thailand

Slični zadaci

Find all pairs of functions f, g : \mathbb R \to \mathbb R such that f \left( x + g(y) \right) = x \cdot f(y) - y \cdot f(x) + g(x) for all real x, y.
Find all functions f: \mathbb{R} \rightarrow \mathbb{R}, satisfying

f(xy)(f(x) - f(y)) = (x-y)f(x)f(y)
for all x,y.
Does there exist a function s\colon \mathbf{Q} \rightarrow \{-1,1\} such that if x and y are distinct rational numbers satisfying {xy=1} or {x+y\in \{0,1\}}, then {s(x)s(y)=-1}? Justify your answer.
Find all functions f: \mathbb{R}\to\mathbb{R} such that f\left(x+y\right)+f\left(x\right)f\left(y\right)=f\left(xy\right)+2xy+1 for all real numbers x and y.
Consider those functions f: \mathbb{N} \mapsto \mathbb{N} which satisfy the condition
f(m + n) \geq f(m) + f(f(n)) - 1
for all m,n \in \mathbb{N}. Find all possible values of f(2007).

Author: unknown author, Bulgaria
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