$9.269 \frac{1}{\log _{1 / 2} \sqrt{x+3}} \leq \frac{1}{\log _{1 / 2}(x+1)}$.

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$9.269 \frac{1}{\log _{1 / 2} \sqrt{x+3}} \leq \frac{1}{\log _{1 / 2}(x+1)}$.
Solution. Since $x>-1$, then $x+3>2, \sqrt{x+3}>1, \log _{1 / 2} \sqrt{x+3}1, \\ x+1 \geq \sqrt{x+3}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>0, \\ (x+2)(x-1) \geq 0\end{array} \Leftrightarrow x \geq 1\right.\right.$.
2) $\left\{\begin{array}{l}0<x+1<1, . \\ x+1 \leq \sqrt{x+3}\end{array} \Leftrightarrow\left\{\begin{array}{l}-1<x<0, \\ (x+2)(x-1) \leq 0\end{array} \Leftrightarrow-1<x<0\right.\right.$.

Answer: $\quad x \in(-1 ; 0) \cup[1 ;+\infty)$.