I checked numerically that $$ \left( \sum_N p_N \dfrac{C_N}{D_N^2} \right) \left( \sum_N p_N D_N \right) \geq \left( \sum_N p_N \dfrac{C_N}{D_N} \right) $$ where $$ \sum_N p_N = 1 \;, \quad 0\leq p_N\leq 1 \;, \quad 0\leq \dfrac{C_N}{D_N} \leq 1 \;, $$ but I'm not able to prove it analytically. Using Cauchy-Schwarz gives the bound $$ \left( \sum_N p_N \dfrac{C_N}{D_N^2} \right) \left( \sum_N p_N D_N \right) \geq \left(\sum_N \sqrt{p_N \dfrac{C_N}{D_N^2} p_N D_N}\right)^2 = \left(\sum_N p_N \sqrt{\dfrac{C_N}{D_N}} \right)^2 $$ which is smaller than $\left( \sum_N p_N \dfrac{C_N}{D_N} \right)$.
Is there a way to prove my first inequality?