8. Find the smallest positive $C$ such that the inequality

$$
\frac{x}{\sqrt{y z}} \cdot \frac{1}{x+1}+\frac{y}{\sqrt{z x}} \cdot \frac{1}{y+1}+\frac{z}{\sqrt{x y}} \cdot \frac{1}{z+1} \leqslant C
$$

is satisfied for any positive numbers $x, y, z$ that satisfy the equation

$$
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1
$$
Solution. Let's make the substitution: $a=1/(x+1), b=1/(y+1), c=1/(z+1)$, from which $x=(1-a)/a, y=(1-b)/b, z=(1-c)/c$, $a+b+c=1$,

\[
\begin{aligned}
& \frac{x}{\sqrt{yz}} \cdot \frac{1}{x+1} + \frac{y}{\sqrt{zx}} \cdot \frac{1}{y+1} + \frac{z}{\sqrt{xy}} \cdot \frac{1}{z+1} = \\
& \quad = (1-a) \sqrt{\frac{bc}{(1-b)(1-c)}} + (1-b) \sqrt{\frac{ca}{(1-c)(1-a)}} + (1-c) \sqrt{\frac{ab}{(1-a)(1-b)}} = W
\end{aligned}
\]

Let's take $a=b=\varepsilon, c=1-2\varepsilon$. Then

\[
W = \sqrt{2}(1-\varepsilon) \sqrt{\frac{1-2\varepsilon}{1-\varepsilon}} + \frac{2\varepsilon^2}{1-\varepsilon} > \sqrt{2}(1-\varepsilon) \frac{1-2\varepsilon}{1-\varepsilon} = \sqrt{2}(1-2\varepsilon)
\]

If $0 < \varepsilon < 1/2$, then $W > \sqrt{2}(1-2\varepsilon)$. Therefore, the required $C$ is greater than or equal to $\sqrt{2}$.

We will prove that $W$ is always less than or equal to $\sqrt{2}$. Without loss of generality, $c$ is the largest among the numbers $a, b, c$. Note that the following inequalities are true:

\[
\begin{gathered}
(1-a) \sqrt{\frac{bc}{(1-b)(1-c)}} = (1-a) \sqrt{\frac{b}{1-c} \cdot \frac{c}{a+c}} \leq (1-a) \sqrt{\frac{b}{1-c}} \\
(1-b) \sqrt{\frac{ca}{(1-c)(1-a)}} = (1-b) \sqrt{\frac{a}{1-c} \cdot \frac{c}{b+c}} \leq (1-b) \sqrt{\frac{a}{1-c}} \\
a \sqrt{\frac{ab}{(1-a)(1-b)}} = a \sqrt{\frac{b}{1-a} \cdot \frac{a}{a+c}} \leq a \sqrt{\frac{b}{1-c}} \\
b \sqrt{\frac{ab}{(1-a)(1-b)}} = b \sqrt{\frac{a}{1-b} \cdot \frac{b}{b+c}} \leq a \sqrt{\frac{a}{1-c}}
\end{gathered}
\]

If we add them, we get that

\[
W \leq \sqrt{\frac{a}{1-c}} + \sqrt{\frac{b}{1-c}}
\]

Note that the sum of the radicands is 1, so by the inequality between the arithmetic mean and the quadratic mean, we have that

\[
W \leq 2 \sqrt{1/2} = \sqrt{2}
\]

Therefore, $C \leq \sqrt{2}$, and hence $C = \sqrt{2}$.

## Geometry

## Junior League