Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[f(x + y) + y \le f(f(f(x)))\] holds for all $x, y \in \mathbb{R}$.

A function $f: R \to R$ satisfies the conditions: $f (x + 19) \le f (x) + 19$ and $f (x + 94) \ge f (x) + 94$ for all $x \in R$. Prove that $f (x + 1) = f (x) + 1$ for all $x \in R$.

Show that for all non-constant functions $f:\mathbb R\to\mathbb R$, there are two real numbers $x,y$ such that $$f(x+f(y))>xf(y)+x.$$

Let $I\subset \mathbb R$ be a non-empty open interval, $\varepsilon\geq 0$ and $f\colon I\rightarrow\mathbb R$ a function satisfying the $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)+\varepsilon t(1-t)|x-y|$$ inequality for all $x,y\in I$ and $t\in [0,1]$. Prove that there exists a convex $g\colon I\rightarrow\mathbb R$ function, such that the function $l :=f-g$ has the $\varepsilon$-Lipschitz property, that is $$|l(x)-l(y)|\leq \varepsilon|x-y|\text{ for all }x,y\in I$$

Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\] for every $x, y \in \mathbb R_{>0}$.

Let $ S\subseteq\mathbb{R}$ be a set of real numbers. We say that a pair $ (f, g)$ of functions from $ S$ into $ S$ is a [i]Spanish Couple[/i] on $ S$, if they satisfy the following conditions: (i) Both functions are strictly increasing, i.e. $ f(x) < f(y)$ and $ g(x) < g(y)$ for all $ x$, $ y\in S$ with $ x < y$; (ii) The inequality $ f\left(g\left(g\left(x\right)\right)\right) < g\left(f\left(x\right)\right)$ holds for all $ x\in S$. Decide whether there exists a Spanish Couple [list][*] on the set $ S \equal{} \mathbb{N}$ of positive integers; [*] on the set $ S \equal{} \{a \minus{} \frac {1}{b}: a, b\in\mathbb{N}\}$[/list] [i]Proposed by Hans Zantema, Netherlands[/i]

Find all functions $f : \mathbb R \to \mathbb R$ for which \[f(a - b) f(c - d) + f(a - d) f(b - c) \leq (a - c) f(b - d),\] for all real numbers $a, b, c$ and $d$. Note that there is only one occurrence of $f$ on the right hand side!

Give examples of a function $f : R \to R$ that satisfies $0 < f(x) < f(x + f(x)) <\sqrt2 x$, for all positive $x$, and show that there is no function $f : R \to R$ that satisfies $x < f(x + f(x)) <\sqrt2 f(x)$, for all positive $x$.

Let $ S\subseteq\mathbb{R}$ be a set of real numbers. We say that a pair $ (f, g)$ of functions from $ S$ into $ S$ is a [i]Spanish Couple[/i] on $ S$, if they satisfy the following conditions: (i) Both functions are strictly increasing, i.e. $ f(x) < f(y)$ and $ g(x) < g(y)$ for all $ x$, $ y\in S$ with $ x < y$; (ii) The inequality $ f\left(g\left(g\left(x\right)\right)\right) < g\left(f\left(x\right)\right)$ holds for all $ x\in S$. Decide whether there exists a Spanish Couple [list][*] on the set $ S \equal{} \mathbb{N}$ of positive integers; [*] on the set $ S \equal{} \{a \minus{} \frac {1}{b}: a, b\in\mathbb{N}\}$[/list] [i]Proposed by Hans Zantema, Netherlands[/i]

Find all functions $f:\mathbb R^+ \to \mathbb R^+$ satisfying \begin{align*} f\left(\frac{1}{x}\right) \geq 1 - \frac{\sqrt{f(x)f\left(\frac{1}{x}\right)}}{x} \geq x^2 f(x), \end{align*} for all positive real numbers $x$.

Consider all polynomials $P(x)$ with real coefficients that have the following property: for any two real numbers $x$ and $y$ one has \[|y^2-P(x)|\le 2|x|\quad\text{if and only if}\quad |x^2-P(y)|\le 2|y|.\] Determine all possible values of $P(0)$. [i]Proposed by Belgium[/i]

Let $ f : [0, 1] \to\ R $ be a function such that :- $1.)$ $f(x) \ge 0$ for all $x$ in $[0,1]$ . $2.)$ $f(1) = 1$ . $3.)$ If $x_1 , x_2$ are in $[0,1]$ such that $x_1 + x_2 \le 1$ , then $f(x_1) + f(x_2) \le f(x_1 + x_2)$ . Show that $f(x) \le 2x $ for all $x$ in $ [0,1] $.

Let $\mathcal{F}$ be the family of bijective increasing functions $f\colon [0,1] \to [0,1]$, and let $a \in (0,1)$. Determine the best constants $m_a$ and $M_a$, such that for all $f \in \mathcal{F}$ we have \[m_a \leq f(a) + f^{-1}(a) \leq M_a.\] [i](Dan Schwarz)[/i]

A function $f : [0,1] \to R$ has the following properties: (a) $f(x) \ge 0$ for $0 < x < 1$, (b) $f(1) = 1$, (c) $f(x+y) \ge f(x)+ f(y) $ whenever $x,y,x+y \in [0,1]$. Prove that $f(x) \le 2x$ for all $x \in [0,1]$.

Let $c > 1$ be a real number. A function $f: [0 ,1 ] \to R$ is called c-friendly if $f(0) = 0, f(1) = 1$ and $|f(x) -f(y)| \le c|x - y|$ for all the numbers $x ,y \in [0,1]$. Find the maximum of the expression $|f(x) - f(y)|$ for all [i]c-friendly[/i] functions $f$ and for all the numbers $x,y \in [0,1]$.

Find all functions $f : Z^+ \to Z^+$ such that $f(k + 1) >f(f(k))$ for $k > 1$, where $Z^+$ is the set of positive integers.

Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that \[f\left(x-f(y)\right)>yf(x)+x\] [i]Proposed by Igor Voronovich, Belarus[/i]

Find all functions $f : R \to R$ such that $$f(x + y) \ge xf(x) + yf(y)$$, for all $x, y \in R$ .

Let $S$ denote the set of $2$ by $2$ matrices with integer entries and determinant $1$, and let $T$ denote those matrices of $S$ which are congruent to the identity matrix $I\pmod3$ (so $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in T$ means that $a,b,c,d\in\mathbb Z,ad-bc=1,$ and $3$ divides $b,c,a-1,d-1$). (a) Let $f:T\to\mathbb R$ be a function such that for every $X,Y\in T$ with $Y\ne I$, either $f(XY)>f(X)$ or $f(XY^{-1})>f(X)$. Show that given two finite nonempty subsets $A,B$ of $T$, there are matrices $a\in A$ and $b\in B$ such that if $a'\in A$, $b'\in B$ and $a'b'=ab$, then $a'=a$ and $b'=b$. (b) Show that there is no $f:S\to\mathbb R$ such that for every $X,Y\in S$ with $Y\ne\pm I$, either $f(XY)>f(X)$ or $f(XY^{-1})>f(X)$.

Given a function $f\colon \mathbb{R}\rightarrow \mathbb{R} $ with $f(x)^2\le f(y)$ for all $x,y\in\mathbb{R} $, $x>y$, prove that $f(x)\in [0,1] $ for all $x\in \mathbb{R}$.

Find all functions $f : Z \to Z$ satisfying the following two conditions: (i) for all integers $x$ we have $f(f(x)) = x$, (ii) for all integers $x$ and y such that $x + y$ is odd, we have $f(x) + f(y) \ge x + y$.

$f(x)$ is a real valued function defined for $x \geq 0$ such that $f(0) = 0$, $f(x+1)=f(x)+\sqrt{x}$ for all $x$, and \[ f(x) < \frac{1}{2}f\left(x - \frac{1}{2}\right)+\frac{1}{2}f\left(x + \frac{1}{2}\right) \quad \text{for all} \quad x \geq \frac{1}{2} \] Show that $f\left(\frac{1}{2}\right)$ is uniquely determined.

Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\] for every $x, y \in \mathbb R_{>0}$.

Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\] for every $x, y \in \mathbb R_{>0}$.

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[f(x + y) + y \le f(f(f(x)))\] holds for all $x, y \in \mathbb{R}$.