3. Let $a, b, c$ be positive real numbers such that $a b + b c + c a = 1$. Determine the values that the expression

$$
\frac{a\left(b^{2}+1\right)}{a+b}+\frac{b\left(c^{2}+1\right)}{b+c}+\frac{c\left(a^{2}+1\right)}{c+a}
$$

can take.

(Josef Tkadlec \& Patrik Bak)
SOLUTION. Given the condition $a b+b c+c a=1$, we first modify the first of the three addends of the expression under consideration, similarly to problem 4 from the homework round,

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

Similarly, the remaining two addends have the expressions

$$
\frac{b\left(c^{2}+1\right)}{b+c}=b c+a b \quad \text{and} \quad \frac{c\left(a^{2}+1\right)}{c+a}=c a+b c
$$

Together, we get for the sum of all three given fractions

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

where we have again used the equality $a b+b c+c a=1$ from the problem statement.

Conclusion. The given expression always takes the value 2 for any positive real numbers $a, b, c$ that satisfy the condition $a b+b c+c a=1$.

For a complete solution to the problem, award 6 points, of which: 3 points for the necessary modification of any of the three addends of the given expression; 2 points for adding the three modified addends (fractions); 1 point for providing the correct answer.

For an unproven hypothesis that the given expression always takes the value 2, award 1 point, even if it is supported by specific numerical calculations.


[^0]:    * These known facts do not need to be proven in the solution. Nevertheless, we note that they follow from the properties of the medians $S K$ and $L S$ in triangles $A B C$ and $A B D$.