10. We shall first prove the inequality for $n$ of the form $2^{k}, k=0,1,2, \ldots$ The case $k=0$ is clear. For $k=1$, we have
$$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}-\frac{2}{\sqrt{r_{1} r_{2}}+1}=\frac{\left(\sqrt{r_{1} r_{2}}-1\right)\left(\sqrt{r_{1}}-\sqrt{r_{2}}\right)^{2}}{\left(r_{1}+1\right)\left(r_{2}+1\right)\left(\sqrt{r_{1} r_{2}}+1\right)} \geq 0$$
For the inductive step it suffices to show that the claim for $k$ and 2 implies that for $k+1$. Indeed,
$$\begin{aligned} \sum_{i=1}^{2^{k+1}} \frac{1}{r_{i}+1} & \geq \frac{2^{k}}{\sqrt[2^{k}]{r_{1} r_{2} \cdots r_{2^{k}}}+1}+\frac{2^{k}}{\sqrt[2^{k}]{r_{2^{k}+1^{\prime} r_{2^{k}}+2^{\cdots r_{2^{k+1}}}+1}}} \\ & \geq \frac{2^{k+1}}{\sqrt[2^{k+1}]{r_{1} r_{2} \cdots r_{2^{k+1}}}+1} \end{aligned}$$
and the induction is complete. We now show that if the statement holds for $2^{k}$, then it holds for every $n$. The function $f(x)=\frac{1}{1+e^{x}}$ is convex for $x>0$ : indeed, $f^{\prime \prime}(x)=\frac{e^{x}\left(e^{x}-1\right)}{\left(e^{x}+1\right)^{3}}>0$. Thus by Jensen's inequality applied to $f\left(x_{1}\right), \ldots, f\left(x_{n}\right)$, we get $\frac{1}{r_{1}+1}+\cdots+\frac{1}{r_{n}+1} \geq \frac{n}{\sqrt[n]{r_{1} \cdots r_{n}}+1}$.