To better see This question you can See here: https://stats.stackexchange.com/questions/321445/how-do-i-show-this-using-the-cauchy-schwarz-inequality
I asked it there. there is a good answer there but it seems not sharp enough.
From Two Population We take two $p$ dimensional samples $\mathbf{X}_1$ and $\mathbf{X}_2$ with sample size $n_1$ and $n_2$ respectively, Let $\mathbf{\Lambda} = \mathrm{diag}\{(\sigma_{11}^2 + \gamma\sigma_{21}^2)^{-1/2},\dots,(\sigma_{1p}^2+ \gamma\sigma^2_{2p})^{-1/2}\}$ which $\gamma$ is $\frac{n_1}{n_2}$, $\boldsymbol{\mu}_1$ and $\boldsymbol{\mu}_2$ are the $p \times 1$ mean vectors and $\Sigma_1$ and $\Sigma_2$ are the $p \times p$ Covariance Matrices of two populations.
if we set $\frac{n_1}{n_1 + n_2} \rightarrow \kappa$ and $\tilde{\boldsymbol{\Sigma}} = (1-\kappa)\boldsymbol{\Sigma}_1 + \kappa\boldsymbol{\Sigma}_2$
Q: How can I show this result using the Cauchy-Schwarz inequality:
$$ \frac{\mathrm{tr}(\boldsymbol{\Lambda}^2)}{p} \sqrt{\frac{\mathrm{tr}(\tilde{\boldsymbol{\Sigma}}^2)}{\mathrm{tr}((\tilde{\boldsymbol{\Lambda}\boldsymbol{\Sigma}\boldsymbol{\Lambda}})^2)}}>1$$
[My Source is this article][1]. Any Help will be appreciated.
[1]: https://www.researchgate.net/publication/279070596_A_note_on_high-dimensional_two-sample_test
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