Moments of a weighted sum of distributions

Let $$X$$ be a random variable with distribution $g(x) = \Sigma_{i=1}^{N}\lambda_{i}f_{i}(x)$, where each $f_{i}$ is a probability distribution for the RV $X_{i}$ (with mean $\mu_{i}$ and variance $\sigma_{i}^2$), $\lambda_{i}\geq0$ and $\Sigma_{i=1}^{N}\lambda_{i}=1$.

We determine the mean, $\mathbb{E}[X]=\mu$ and variance, $\mathbb{V}[X]=\sigma^2$, of $X$ in terms of those of the $X_{i}$.

We have
$$\mu =    \mathbb{E}[X] =    \mathbb{E}[\Sigma_{i=1}^{N}\lambda_{i}X_{i}] =    \Sigma\lambda_{i}\mathbb{E}[X_{i}] =    \Sigma_{i=1}^{N}\lambda_{i}\mu_{i}$$
by linearity.

Then on expanding the square we have
$$\sigma^2 = \mathbb{V}[X]=\mathbb{E}[(X-\mu)^{2}]=\mathbb{E}[X^{2}]-\mu^{2}$$ ,
but that isn't really any help to us, because we haven't determined $\mathbb{E}[X^2]$.

Alternatively,
$$\sigma^{2}   = \int dx(x-\mu)^{2}f(x)   = \int dx(x-\mu)^{2}\Sigma_{i}\lambda_{i}f_{i}(x)   = \Sigma_{i}\lambda_{i}\int(x-\mu)^{2}f_{i}(x)$$
$$= \Sigma_{i}\lambda_{i}\int dxf_{i}(x)(x-\mu_{i}+\mu_{i}-\mu)^{2}$$
$$= \Sigma_{i}\lambda_{i}\int dxf_{i}(x)[(x-\mu_{i})^{2}+2(x-\mu_{i})(\mu_{i}-\mu)+(\mu_{i}-\mu)^{2}]$$

$$=\Sigma_{i}\lambda_{i}[\int dxf_{i}(x)(x-\mu_{i})^{2}+2(\mu_{i}-\mu)\int dxf_{i}(x)(x-\mu_{i})+(\mu_{i}-\mu)^{2}\int dxf_{i}(x)]$$

$$= \Sigma_{i=1}^{N}\lambda_{i}(\mathbb{E}[(X_{i}-\mu_{i})^{2}]+2(\mu_{i}-\mu)(\mathbb{E}[X_{i}]-\mu_{i})+(\mu_{i}-\mu)^{2})$$

$$= \Sigma_{i=1}^{N}\lambda_{i}(\mathbb{V}[X_{i}]+(\mu_{i}-\mu)^{2})$$
$$\geq \Sigma_{i} \lambda_{i} \mathbb{V}[X_{i}]$$