Proving a triangle inequality for some divergence The variational distance is defined by,
$$
V(P,Q)=\sum _{i}|p_{i} -q_{i} | 
$$ 
where $P=(p_{1} ,...,p_{n})$ and $Q=(q_{1} ,...,q_{n} )$ are discrete distributions.

It is fairly easy to see that $V$ is a metric, and in particular that it satisfies the triangle inequality. I now introduce class prior, $0&lt;\alpha&lt;1$, which represents the relative 'size' of the distributions,
$$
V(\alpha ,P,Q)=\sum _{i}|\alpha p_{i} -(1-\alpha )q_{i} |
$$
and would like to prove that it again satisfies the triangle inequality,

$$V(\alpha ,P,Q)+V(\beta ,Q,R)\ge V(\gamma ,P,R) $$
where $R$ is another distribution, $\beta$ and $\gamma$ are respective priors, and $\gamma$ is naturally, 
$$\gamma =\frac{\alpha \beta }{\alpha \beta +(1-\alpha )(1-\beta )} 
$$

I have confidence (numerically) that this triangle inequality indeed holds, but would like a formal proof. Any ideas?
Actually, I've recently realized that the inequality is false. It is true only for equal class priors. Sorry about that.
But $V(\alpha, P, Q) != V(\alpha, Q, P)$ in general, therefore it would not be a metric, right? If that is true, what is the point of proving or disproving the triangle inequality?