## Task A-1.1.

Depending on the real parameter $m$, determine for which real numbers $x$ the following inequality holds:

$$
\frac{x-m}{x^{2}}+x \geqslant 2\left(1-\frac{m}{x}\right)+m
$$
## Solution.

Note that $x$ must be different from zero for all expressions to make sense. 1 point

By moving all addends to the same side, we have

$$
\frac{x-m}{x^{2}}-2 \cdot \frac{x-m}{x}+x-m \geqslant 0
$$

1 point

Further simplification yields

$$
(x-m)\left(\frac{1}{x^{2}}-\frac{2}{x}+1\right) \geqslant 0
$$

or

$$
(x-m)\left(\frac{1}{x}-1\right)^{2} \geqslant 0
$$

Since $\left(\frac{1}{x}-1\right)^{2} \geqslant 0$ always holds, it follows that $x \geqslant m$ or $x=1$.

If $m>1$, the solution is $x \in\{1\} \cup[m,+\infty\rangle$. 1 point

If $0 \leqslant m \leqslant 1$, the solution is $x \in[m,+\infty\rangle$. 1 point

If $m<0$, the solution is $x \in[m,+\infty\rangle \backslash\{0\}$. 1 point

Note: If the student does not exclude the possibility of $x=0$ or does not include the possibility of $x=1$, they lose 2 points for each of these cases.