I have two finite probability mass functions (pmfs) $P(x)$ and $Q(x)$ on the same support $(0,1,\ldots,n)$. Let $(p_0,p_1,\ldots,p_n)$ and $(q_0,q_1,\ldots,q_n)$ be the probability vectors from the two functions, $0<p_i<1$, $0<q_j<1$, $\sum_i^np_i=1$, $\sum_j^nq_j=1$.
It is known that the mean of $P(x)$ is less than that of $Q(x)$, and the coefficient of variation (cv) of $P(x)$ is greater than that of $Q(x)$, namely,
$$\sum_t^ntp_t < \sum_t^ntq_t,$$ and $$\frac{\sqrt{\sum_t^nt^2p_t-(\sum_t^ntp_t)^2 }}{\sum_t^ntp_t} > \frac{\sqrt{\sum_t^nt^2q_t-(\sum_t^ntq_t)^2 }}{\sum_t^ntq_t}\text{, or simply, } \frac{\sum_t^nt^2p_t}{(\sum_t^ntp_t)^2} > \frac{\sum_t^nt^2q_t}{(\sum_t^ntq_t)^2} .$$
Now I want to prove that the cv of a linear mixture of the two pmfs is between the two pmfs' cvs, namely, for $0<w<1$,
$$\frac{\sum_t^nt^2p_t}{(\sum_t^ntp_t)^2} > \frac{ (1-w) \sum_t^nt^2p_t + w\sum_t^nt^2q_t }{ \left[ (1-w) \sum_t^ntp_t + w\sum_t^ntq_t\right]^2 } > \frac{\sum_t^nt^2q_t}{(\sum_t^ntq_t)^2} .$$
There are two other pieces of information that I am not sure if they are crucial to the proof, namely, $$p_0>q_0\text{ and }p_n<q_n.$$
Massive simulations show the inequality seems to hold. I sense the key to proving it is to transform the inequality as a function of $w$, but the math becomes messy quite soon.
Proving the inequality is vital to showing the correctness of an algorithm I am trying to make. Is the inequality really true or does there exist a counter example?
Thank you so much in advance!