Is excess kurtosis for a mixture of two normal distributions with the same means and different variances always positive? For a mixture of two normal distributions with the same mean the excess kurtosis is stated to be $$3 \frac{(p_1 \sigma_1^4 + p_2 \sigma_2^4)}{(p_1\sigma_1^2 + p_2\sigma_2^2)^2} -3$$ I believe that this should always be positive when the sigmas are always > 0 and the p's are from (0,1). This website implies that the demonstration relies on the relationship between arithmetic and geometric weighted sums, but I don't see it. Is the claim true? And if so, can you help me see how to demonstrate it.

You forgot to mention $p_1 + p_2 = 1$ which follows from the web site you linked. I will use $s,t$ for $\sigma_1, \sigma_2$ and $p,q$ for $p_1, p_2 = 1-p_1$ to simplify notation.

Sufficient to show that $$ps^4 + qt^4 \geq \left(ps^2 + qt^2\right)^2.$$

Note that $$ \begin{split} 0 &\leq \left(s^2 - t^2\right)^2 = s^4+t^4 - 2s^2 t^2 \\\ 2s^2 t^2 &\leq s^4 + t^4\\\ 2pq s^2 t^2 &\leq s^4pq + t^4pq = s^4p(1-p) + t^4q(1-q)\\\ 2pq s^2 t^2 &\leq s^4p+ t^4q -s^4p^2 - t^4 q^2 \\\ 2pq s^2 t^2 + s^4p^2 + t^4 q^2 &\leq s^4p+ t^4q \\\ \left(ps^2 + qt^2\right)^2 &\leq s^4p+ t^4q \\\ \end{split} $$ as desired.