1. For non-negative real numbers $a, b$, it holds that $a+b=2$. Determine the smallest and largest possible value of the expression

$$
V=\frac{a^{2}+b^{2}}{a b+1}
$$

(Patrik Bak)
Solution. Let's modify the expression $V$ in the following way:

$$
V=\frac{a^{2}+b^{2}}{a b+1}=\frac{(a+b)^{2}-2 a b}{a b+1}=\frac{4-2 a b}{a b+1}=1+\frac{3(1-a b)}{a b+1} .
$$

Thanks to the fact that $a b \geq 0$, from the second-to-last expression of $V$ we get the estimate

$$
V \leq \frac{4-0}{0+1}=4
$$

with equality $V=4$ being achieved precisely when $a b=0$, which is satisfied by the two admissible pairs $(a, b)=(2,0)$ and $(a, b)=(0,2)$.

From the well-known inequality $\sqrt{a b} \leq \frac{1}{2}(a+b)$ between the arithmetic and geometric means of two non-negative numbers $a, b$ and the condition $a+b=2$, it follows that $\sqrt{a b} \leq 1$, and thus also $a b \leq 1$. Therefore, the fraction in the last expression (1) is non-negative, and hence $V \geq 1$. Equality $V=1$ occurs precisely when $a b=1$, which, according to our derived inequality $a b \leq 1$, is satisfied by the single admissible pair $(a, b)=(1,1)$, since only in the case $a=b$ do both means used equal each other.

Answer. Under the given conditions, the expression $V$ has the smallest value of 1 and the largest value of 4.