4. Let $a, b$ be any positive real numbers such that $a^{2}+b^{2}=1$. Find the smallest possible value of the expression

$$
\frac{a^{2}\left(a+b^{3}\right)}{b-b^{3}}+\frac{b^{2}\left(b+a^{3}\right)}{a-a^{3}}
$$

and determine for which pairs $(a, b)$ this value is achieved. (Tomáš Bárta)
SOLUTION. Using the condition $a^{2}+b^{2}=1$, we first simplify the first fraction of the given expression:

$$
\frac{a^{2}\left(a+b^{3}\right)}{b-b^{3}}=\frac{a^{2}\left(a+b^{3}\right)}{b\left(1-b^{2}\right)}=\frac{a^{2}\left(a+b^{3}\right)}{b \cdot a^{2}}=\frac{a+b^{3}}{b}=\frac{a}{b}+b^{2} .
$$

Similarly, for the second fraction we have

$$
\frac{b^{2}\left(b+a^{3}\right)}{a-a^{3}}=\frac{b}{a}+a^{2}
$$

By adding the expressions of both fractions, we get for the given expression $V$:

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

where we have again used the condition $a^{2}+b^{2}=1$.

Notice that for the sum of the fractions in the last parentheses, we have

$$
\frac{a}{b}+\frac{b}{a}=\frac{a^{2}+b^{2}}{a b}=\frac{(a-b)^{2}+2 a b}{a b}=\frac{(a-b)^{2}}{a b}+2 \geqq 2 \text {, since } \quad \frac{(a-b)^{2}}{a b} \geqq 0 \text {. }
$$

From the derived expression for $V$, it follows that the lower bound is $V \geqq 2+1=3$. Moreover, in the used inequality $a / b+b / a \geqq 2$, equality holds if and only if $a=b$. Therefore, the equality $V=3$ holds if and only if the positive numbers $a, b$ satisfy both conditions $a^{2}+b^{2}=1$ and $a=b$, i.e., if and only if $a=b=\frac{1}{2} \sqrt{2}$. The number 3 is thus the smallest value of the given expression $V$ and is achieved for the unique pair of numbers $a, b$ (determined at the end of the previous sentence).

For a complete solution, award 6 points, of which: 2 points for the necessary simplification of at least one of the two given fractions; 1 point for simplifying their sum to a form suitable for minimizing the sum $a / b+b / a$ or maximizing the product $a b$; 2 points for justifying the lower bound of 3 (one can refer to the AM-GM inequality as in the footnote); 1 point for determining (not just guessing) when the minimum value is achieved.

For an incomplete solution, award 1 point for guessing the minimum value 3 only if the corresponding pair $a=b=\frac{1}{2} \sqrt{2}$ is provided. This point can be added to the first 2 or 3 points awarded according to the instructions in the previous paragraph. Without a complete proof of the inequality $V \geqq 3$, even with other methods, no more than 4 points can be awarded.[^1]

[^0]: * The first step in such a justification might be the consideration of the pair of segments $B C$ and $F G$ from the introductory paragraph of the solution.

[^1]: * The proof of this result can also use the AM-GM inequality $\frac{1}{2}(u+v) \geqq \sqrt{u v}$ for values $u=a / b$ and $v=b / a$. It is also possible to use the equality $a / b+b / a=1 /(a b)$ due to the condition $a^{2}+b^{2}=1$ and then maximize the product $a b$, for example, by using the mentioned AM-GM inequality for $u=a^{2}$ and $v=b^{2}$.