9.9. Find the largest number $m$ such that for any positive numbers $a, b$, and $c$, the sum of which is 1, the inequality

$$
\sqrt{\frac{a b}{c+a b}}+\sqrt{\frac{b c}{a+b c}}+\sqrt{\frac{c a}{b+c a}} \geqslant m
$$

holds.

(l. Emelyanov)
Answer. $m=1$.

First solution. First, we prove that $m=1$ satisfies the requirements of the problem. Notice that $ab + c = ab + c(a + b + c) = (c + a)(c + b)$. Therefore,

\[
\begin{aligned}
& \sqrt{\frac{ab}{c + ab}} + \sqrt{\frac{bc}{a + bc}} + \sqrt{\frac{ca}{b + ca}} = \\
& = \sqrt{\frac{ab}{(c + a)(c + b)}} + \sqrt{\frac{bc}{(a + b)(a + c)}} + \sqrt{\frac{ca}{(b + c)(b + a)}} = \\
& = \frac{\sqrt{ab} \sqrt{a + b} + \sqrt{bc} \sqrt{b + c} + \sqrt{ca} \sqrt{c + a}}{\sqrt{(a + b)(b + c)(c + a)}}.
\end{aligned}
\]

Thus, it remains to prove the inequality

\[
\sqrt{ab} \sqrt{a + b} + \sqrt{bc} \sqrt{b + c} + \sqrt{ca} \sqrt{c + a} \geq \sqrt{(a + b)(b + c)(c + a)}
\]

Square this inequality; it will take the form

\[
\begin{gathered}
ab(a + b) + bc(b + c) + ca(c + a) + 2 \sqrt{ab^2 c(a + b)(b + c)} + \\
+ 2 \sqrt{bc^2 a(b + c)(c + a)} + 2 \sqrt{ca^2 b(c + a)(a + b)} \geq \\
\geq a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2 + 2abc
\end{gathered}
\]

After simplification, the left side will remain the sum of the square roots, and the right side will be $2abc$. But any of the square roots is not less than $abc$; indeed, for example, $\sqrt{ab^2 c(a + b)(b + c)} \geq \sqrt{ab^2 c \cdot ac} = abc$. This is where the required inequality follows.

It remains to prove that for any $m > 1$, the inequality is not always satisfied; it is sufficient to show this for $1a \geq \frac{1}{3}$, so the inequality is not always satisfied - 2 points.

Proving only that $m=1$ satisfies the requirements of the problem - 5 points.