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Let $\mathbb{R}^+$ be the set of positive real numbers. Let $f\colon \mathbb{R}^{+}\to \mathbb{R}^{+}$ be a function such that for all $x,y\in \mathbb{R}^{+}$ it holds that

$$y f^{2025}(x) \geq x f(y).$$

Show that there exists a positive integer $n_0$ such that for all positive integers $n \geq n_0$ and for all $x \in \mathbb{R}^+$ it holds that

$$f^n(x) \geq x.$$

Remark. Here $f^n$ denotes the function $f$ applied $n$ times, this means $f^n(x) = \underbrace{f(f(\ldots f(x)\ldots))}_{n \text{ times}}$.

Let $\mathbb{R}^+$ be the set of positive real numbers. Determine all functions $f\colon \mathbb{R}^{+}\to \mathbb{R}^{+}$ such that for all numbers $x,y\in \mathbb{R}^{+}$, we have $$f(xy) + f(x) = f(y)f(xf(y)) + f(x)f(y),$$

and there exists at most one number $a \in \mathbb{R}^+$ such that $f(a) = 1$.

Odredi sve funkcije $f: \mathbb{R} \to \mathbb{R}$ takve da za sve $x, y \in \mathbb{R}$ vrijedi $$f(f(x) - y^2) = yf(x^2).$$

Odredi sve funkcije $f: \mathbb{R} \to \mathbb{R}$ takve da za sve $x, y \in \mathbb{R}$ vrijedi $$f(f(x)) + f(y) = 2y + f(x - y).$$