How do I show this using the Cauchy-Schwarz inequality From Two Population We have two $p$ dimensional sampels $\mathbf{X}_1$ and $\mathbf{X}_2$ with sample size $n_1$ and $n_2$ respectively, Let We have $\mathbf{\Lambda} = \mathrm{diag}\{(\sigma_{11}^2 + \gamma\sigma_{21}^2)^{-1/2},\dots,(\sigma_{1p}^2+ \gamma\sigma^2_{2p})^{-1/2}\}$ 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 $p \times p$ Variance-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. Any Help will be appreciated.
 A: I am able to use Cauchy-Schwartz inequality, but I not quite getting the same result. I may have made a mistake, so here are all the steps.

Note that for positive semi-definite matrices, trace defines an inner product. That is $tr(AB) = \langle B^T, A \rangle$.
Then by Cauchy-Schwarz for $A$ and $B$ positive semi-definite symmetric matrices,
$$tr(AB) = \langle B, A \rangle  \leq  \sqrt{\langle B,B\rangle  \langle A,A \rangle } = \sqrt{tr(B^2) tr(A^2)} \,.$$
In addition, for positive semi-definite matrices, all eigenvalues are non-negative, so $tr(B^2) \leq tr(B)^2$. Also, the product of positive semi-definite matrices is also a positive semidefinite  if the product is symmetric (so $\Lambda \tilde{\Sigma} \Lambda$ is PSD). Finally, for matrices $A,B,C$, $tr(ABC) = tr(CAB)$. Putting all of this together.
\begin{align*}
\dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{tr((\Lambda \tilde{\Sigma} \Lambda)^2)}}& \geq \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{tr(\Lambda \tilde{\Sigma} \Lambda)^2}}\\
 &= \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{tr(\Lambda^2 \tilde{\Sigma})^2}}\\
& = \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{\langle \tilde{\Sigma}, \Lambda^2\rangle^2}}\\
&\geq \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{ tr(\tilde{\Sigma}^2)tr(\Lambda^4)}}\\
& \geq \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{1}{tr(\Lambda^2)^2}}\\
& = \dfrac{1}{p} 
\end{align*}
I don't get that this is greater than 1. 
