$k <=1$ the example is $g (x)=1/x$ in fact, every decreasing function... if $k>1$, one can prove by induction that $g (x)>=k^{n} g (x+(1+1/k +1/k^{2}+...+ 1/k^{n-1})g (x)) $ and so $g (x+g (x)k/k-1)<=g (x)/k^{n} $ for every natural $n $ absurd!

aleksam wrote: one can prove by induction that $g (x)>=k^{n} g (x+(1+1/k +1/k^{2}+...+ 1/k^{n-1})g (x)) $ Could you explain it, please ?

Put $x \rightarrow x+g(x)$ several times and each step use the inequality $g(x+g(x)) \le \frac{g(x)}{k}$

A problem proposed by Uzbekistan.

Let $k\le 1$. Then $kg(x+g(x)\le g(x+g(x)\le g(x)$ as g(x)>0 Suppose $k>1$. Let $g(x) \le \frac{x}{k}$ for some x. Then $\frac{x}{k}\geq g(x)\geq kg(x+g(x)\geq kg(x+\frac{x}{k})\geq kg(2x)))$. So $g(2x)\le \frac{x}{k^2}$. Putting $x= \frac{x}{2}$ $g(x)\le \frac{x}{2k^2}$. $==>$ $\frac{x}{2k^2}\geq kg(x+\frac{x}{2k^2})\geq kg(\frac{3}{2}x)$. Putting $x=\frac{2}{3}x$ we get $g(x)\le \frac{x}{3k^3}$. And so $g(x)\le \frac{x}{nk^n}$ Then $g(x)\geq kg(\frac{n+1}{n}x)$, where n ---->$+\infty$ and $\frac{n+1}{n}$---->1, so $g(x)\geq kg(x)$ and $k\le 1$. Contradiction.

chessmaster4 wrote: Let $k\le 1$. Then $kg(x+g(x)\le g(x+g(x)\le g(x)$ as g(x)>0 Suppose $k>1$. Let $g(x) \le \frac{x}{k}$ for some x. Then $\frac{x}{k}\geq g(x)\geq kg(x+g(x)\geq kg(x+\frac{x}{k})\geq kg(2x)))$. So $g(2x)\le \frac{x}{k^2}$. Putting $x= \frac{x}{2}$ $g(x)\le \frac{x}{2k^2}$. $==>$ $\frac{x}{2k^2}\geq kg(x+\frac{x}{2k^2})\geq kg(\frac{3}{2}x)$. Putting $x=\frac{2}{3}x$ we get $g(x)\le \frac{x}{3k^3}$. And so $g(x)\le \frac{x}{nk^n}$ Then $g(x)\geq kg(\frac{n+1}{n}x)$, where n ---->$+\infty$ and $\frac{n+1}{n}$---->1, so $g(x)\geq kg(x)$ and $k\le 1$. Contradiction. Why such $x$ that $\frac{x}{k} \geq g(x)$ exists? And you haven’t proven that g is continuous, thus you can’t say if $x -> y$, then $g(x) -> g(y)$

k=1 and k<1