## Task 1.

Determine all pairs $(a, b)$ of real (not necessarily positive) numbers such that $a^{2}+b^{2}=25$ and

$$
a b + a + b
$$

takes the smallest possible value.
## Solution.

Transforming the inequality $(a+b+1)^{2} \geqslant 0$ we get $a^{2}+b^{2}+1+2 a b+2 a+2 b \geqslant 0$, from which it follows that $2(a b+a+b) \geqslant-\left(a^{2}+b^{2}\right)-1$, i.e., $a b+a+b \geqslant-13$.

Equality is achieved if and only if $a+b+1=0, \text{i.e., } b=-a-1$. Substituting into $a^{2}+b^{2}=25$ we get

$$
\begin{aligned}
a^{2}+(-a-1)^{2} & =25 \\
2 a^{2}+2 a+1 & =25 \\
a^{2}+a-12 & =0 \\
(a+4)(a-3) & =0
\end{aligned}
$$

from which the two symmetric solutions follow: $(a, b)=(-4,3)$ and $(a, b)=(3,-4)$.