Draw $C_1$, $C_2$, ... $C_n$ from some distribution, for any $\lambda_i>0$, we hope to get \begin{equation} \underset {C_1,C_2,\cdots, C_n}{\mathbb{E} } \left[ \sqrt{\lambda_1^2 C_1+ \lambda_2^2 C_2 + \cdots + \lambda_n^2 C_n} \right] = \Theta(\lambda_1+ \lambda_2 + \cdots + \lambda_n) \end{equation}
Are there distribution making the above inequality holds?
Assuming that the distribution is supported on $[0,\eta]$, the expected value is less than $$\sqrt{\eta}\sqrt{\lambda_1^2+\ldots+\lambda_n^2} \leq \sqrt{\eta}\,(\lambda_1+\ldots+\lambda_n).\tag{1}$$ A similar inequality holds if the distribution is supported on $\mathbb{R}^+$ and has a finite expected value, in virtue of Jensen's inequality. The converse inequality is more delicate. For sure the Cauchy-Schwarz inequality gives: $$\sqrt{n}\sqrt{\lambda_1^2 C_1+\ldots+\lambda_n^2 C_n}\geq \lambda_1\sqrt{C_1}+\ldots+\lambda_n\sqrt{C_n},\tag{2}$$ hence a sufficient condition for the converse inequality to hold is that our distribution is supported on $\mathbb{R}^+$ and has a finite expected value for $\sqrt{x}$.