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One distance or discrimination of two probability distributions

$$ \textit{$P=(p_{1} ,...,p_{n} )$}$$ and $$\textit{$Q=(q_{1} ,...,q_{n} )$}$$ is the Variational distance defined by, $$ V(P,Q)=\sum _{i}|p_{i} -q_{i} | $$ It is fairly easy to see that V is a metric, and in particular that it satisfies the triangle inequality. I would like to introduce class priors, and define V', $$ V'(\alpha ,P,Q)=\sum _{i}|\alpha p_{i} -(1-\alpha )q_{i} | $$ and prove that V' (while not strictly a metric since the zero is not satisfied) satifies the triangle inequality,

$$V'(\alpha ,P,Q)+V'(\beta ,Q,R)\ge V'(\gamma ,P,R) $$ where $$\gamma =\frac{\alpha \beta }{\alpha \beta +(1-\alpha )(1-\beta )} $$

Any ideas?

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