For positive numbers $a, b, c$ such that $a^{2}+b^{2}+c^{2}=1$. Determine the smallest possible value of

$$
S=\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}
$$
I. solution. According to the condition, $S$ is positive, so the minimum of $S^2$ and $S$ exist simultaneously. The minimum of $S$ is the square root of the minimum of $S^2$, and the two minima are obtained for the same values of the variables. Let's calculate $S^2$:

$$
S^{2}=\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}}+2(\underbrace{\frac{a b}{c} \cdot \frac{a c}{b}+\frac{a b}{c} \cdot \frac{b c}{a}+\frac{b c}{a} \cdot \frac{a c}{b}}_{a^{2}+b^{2}+c^{2}=1})
$$

Rearrange the obtained equality and group the terms on the right side:

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

The right side, according to the arithmetic and geometric mean inequality, is greater than or equal to

$$
\sqrt{\frac{a^{2} b^{2}}{c^{2}} \cdot \frac{a^{2} c^{2}}{b^{2}}}+\sqrt{\frac{a^{2} b^{2}}{c^{2}} \cdot \frac{b^{2} c^{2}}{a^{2}}}+\sqrt{\frac{b^{2} c^{2}}{a^{2}} \cdot \frac{a^{2} c^{2}}{b^{2}}}=a^{2}+b^{2}+c^{2}=1
$$

We obtained that $S^{2} \geq 3$; therefore, according to the introduction, $S$ has a minimum value under the given conditions, and that value is $\sqrt{3}$. It is also clear that equality holds if and only if the equality holds in all three arithmetic-geometric mean inequalities, which happens precisely when $\frac{a b}{c}=\frac{a c}{b}=\frac{b c}{a}$, i.e., $a=b=c=\frac{\sqrt{3}}{3}$.

Remarks. 1. The reader may recognize the method of the Muirhead theorem presented in our October 2005 issue. The essence of the problem was to prove the inequality

$$
\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}} \geq a^{2}+b^{2}+c^{2}
$$

This is a symmetric version of the Muirhead theorem, which is true since $(2 ; 2 ;-2) \succ(2 ; 0 ; 0)$. The statement of the problem is sharper, as the inequality holds even among cyclic permutations. This is due to the fact that the point $B(2 ; 0 ; 0)$ lies within the convex hull of the points $A_{1}(2 ; 2 ;-2), A_{2}(2 ;-2 ; 2), A_{3}(-2 ; 2 ; 2)$: the midpoint of the segment $A_{1} A_{2}$.

2. We can also prove the key inequality in the above solution,

$$
Q=\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}} \geq a^{2}+b^{2}+c^{2}
$$

by grouping the terms in the double of the left side appropriately:

$$
\begin{aligned}
& \left(\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}}\right)+\left(\frac{c^{2} a^{2}}{b^{2}}+\frac{a^{2} b^{2}}{c^{2}}\right)+\left(\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}}\right)= \\
& =b^{2}\left(\frac{a^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\right)+a^{2}\left(\frac{c^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}\right)+c^{2}\left(\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}\right)
\end{aligned}
$$

On the right side, each parenthesis contains the sum of a positive number and its reciprocal, which is known to be at least 2. Thus, $2 Q \geq 2\left(a^{2}+b^{2}+c^{2}\right)$, which is what we wanted to prove.

II. solution. We will prove that $S \geq \sqrt{3}$. After multiplying by the positive denominators, it is sufficient to prove that $\sqrt{3} a b c \leq a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}$.

Introduce the variables $x=a b, y=b c, z=c a$. With these, the squares of the original variables are

$$
a^{2}=\frac{z x}{y}, \quad b^{2}=\frac{x y}{z}, \quad c^{2}=\frac{y z}{x}
$$

The condition, after multiplying by the denominators, in terms of the new variables means that $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}=x y z$, and the statement to be proved is that

$$
\sqrt{3 x y z} \leq x^{2}+y^{2}+z^{2}
$$

Squaring both sides,

$$
3 x y z \leq x^{4}+y^{4}+z^{4}+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)
$$

Substitute the square sum form of $x y z$ from the condition for the new variables:

$$
3 x^{2} y^{2}+3 y^{2} z^{2}+3 z^{2} x^{2} \leq x^{4}+y^{4}+z^{4}+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)
$$

Rearranging and multiplying by 2,

$$
\begin{aligned}
0 & \leq x^{4}+y^{4}-2 x^{2} y^{2}+y^{4}+z^{4}-2 y^{2} z^{2}+z^{4}+x^{4}-2 z^{2} x^{2}= \\
& =\left(x^{2}-y^{2}\right)^{2}+\left(y^{2}-z^{2}\right)^{2}+\left(z^{2}-x^{2}\right)^{2}
\end{aligned}
$$

The final inequality is clearly true. This proves that $S \geq \sqrt{3}$. Equality is possible precisely when $x^{2}=y^{2}=z^{2}$, i.e., when $a^{2}=b^{2}=c^{2}=\frac{1}{3}$, so the smallest possible value of the expression is $\sqrt{3}$.