## Task 3A - 201233A

All natural numbers $n$ are to be determined that have the following property:

For all real numbers $a$ and $b$ with $0<a<b$ it holds that

$$
a+\frac{1}{1+a^{n}}<b+\frac{1}{1+b^{n}}
$$
The task is equivalent to the question of whether $f_{n}(x):=x+\frac{1}{1+x^{n}}$ is strictly monotonically increasing for $x>0$. It holds that

$$
f_{n}^{\prime}(x):=1-\frac{n x^{n-1}}{\left(1+x^{n}\right)^{2}} ; \quad f_{n}^{\prime}(1):=\frac{4-n}{4}
$$

Therefore, for $n>4$, the function is strictly monotonically decreasing at $x=1$. For $n=4$, it follows from $f_{4}^{\prime \prime}(1)>0$ that it has a local minimum there.

We now show that $f_{n}$ is strictly monotonically increasing for $n \leq 3$ and $x>0$.

$$
f_{n}^{\prime}(x)=\frac{\left(1+x^{n}\right)^{2}-n x^{n-1}}{\left(1+x^{n}\right)^{2}}=\frac{\left(1-x^{n}\right)^{2}+x^{n-1}(4 x-n)}{\left(1+x^{n}\right)^{2}}
$$

The first summand in the numerator and the denominator are always non-negative. The second summand is positive for $x \geq 1>\frac{n}{4}$. Thus, $f_{n}$ is strictly monotonically increasing for $x \geq 1$. For $0<x<1$, the second summand $x^{n-1}(4x-n)$ is also positive.

## Source anonymous