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

$$
V=\frac{a^{4}+b^{4}+a b+1}{a+b} .
$$

(Patrik Bak, Jaromír Šimša)
SOLUTION. First, note that the expression $V$ always makes sense under the given conditions on the numbers $a, b$, because the equality $a^{2}+b^{2}=1$ excludes the possibility $a=b=0$, and thus the sum of non-negative numbers $a, b$ is different from zero (i.e., positive).

Determination of the minimum. From the generally valid inequality $(a-b)^{2} \geqq 0$, in our situation we have $2 a b \leqq a^{2}+b^{2}=1$, and thus $0 \leqq a b \leqq 1 / 2$. In addition, the inequality $2 a b \leqq 1$ together with the equality $(a+b)^{2}=1+2 a b$ leads to the estimates $1 \leqq(a+b)^{2} \leqq 2$, from which, after taking the square root, we get $1 \leqq a+b \leqq \sqrt{2}$. Therefore, we have

$$
V=\frac{\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}+a b+1}{a+b}=\frac{2+a b(1-2 a b)}{a+b} \geqq \frac{2}{a+b} \geqq \frac{2}{\sqrt{2}}=\sqrt{2},
$$

and the equality $V=\sqrt{2}$, as can be seen, occurs precisely when $2 a b=1$ (i.e., $a=b$, as follows from the initial sentence of this paragraph) and simultaneously $a+b=\sqrt{2}$, i.e., only for $a=b=\frac{1}{2} \sqrt{2}$.

Determination of the maximum. Due to the assumption $a^{2}+b^{2}=1$, we have

$$
\begin{aligned}
V & =\frac{(a+b)\left(a^{3}+b^{3}\right)-a b\left(a^{2}+b^{2}\right)+a b+1}{a+b}=\frac{(a+b)\left(a^{3}+b^{3}\right)+1}{a+b}= \\
& =a^{3}+b^{3}+\frac{1}{a+b} \leqq 1+\frac{1}{1}=2,
\end{aligned}
$$

since, due to the equality $a^{2}+b^{2}=1$, it must hold that $0 \leqq a \leqq 1$ and $0 \leqq b \leqq 1$, from which the estimates $a^{3} \leqq a^{2} \leqq a$, resp. $b^{3} \leqq b^{2} \leqq b$, follow, and after adding them we get

$$
a^{3}+b^{3} \leqq a^{2}+b^{2} \leqq a+b \quad \text { or } \quad a^{3}+b^{3} \leqq 1 \leqq a+b \text {. }
$$

The equality $V=2$ occurs precisely when both sums $a^{3}+b^{3}, a+b$ are (as well as the sum $a^{2}+b^{2}$) equal to 1, i.e., in the cases where $\{a, b\}=\{0,1\}$.

Answer. The minimum value of the expression $V$ is $\sqrt{2}$, its maximum value is 2.