Let ${\bf \Delta}=(\delta_1,\,\ldots,\,\delta_n)\in\mathbb{R}^n\setminus(0,\,\ldots,\,0)$, ${\bf 1}=(1,\,\ldots,\,1)_{1\times n}$ and $\bf 1\Delta$$^T=0$, where $T$ denotes transpose.
Let ${\bf \mu}=(\mu_1-r_f,\,\ldots,\,\mu_n-r_f)$, where $\mu_1,\,\ldots,\,\mu_n$ are expected returns of some random returns $r_1,\,\ldots,\,r_n$ and $r_f$ denotes the risk-free return rate.
Let $\Sigma$ denote the covariance matrix of $r_1,\,\ldots,\,r_n$ and $\Sigma^{-1}$ its inverse. We assume that $\Sigma^{-1}$ exists.
Can you show that $$ \left({\bf \Delta \mu}^T\right)^2<{\bf \mu}\Sigma^{-1}{\bf \mu}^T\,{\bf \Delta}\Sigma{\bf \Delta}^T $$ for all such $\bf \Delta$?
Such inequality appears trying to justify that
$$ {\bf W}^T=(w_1,\,\ldots,\,w_n)^T=\frac{\Sigma^{-1}{\bf \mu}^T}{{\bf 1}\Sigma^{-1}{\bf \mu}^T} $$ maximizes the Sharpe ratio
$$ \frac{\bf W \mu^T}{\sqrt{{\bf W}\Sigma{\bf W}^T}},\text{ when } {\bf 1}{\bf W}^T=1. $$
Since the covariance matrix of non-degenarate random variables is positively defined, then $$ \left({\bf 1}\Sigma^{-1}{\bf\mu}^T\right)^2{\bf W}\Sigma\bf{W}^T>0 $$ for any collection of weights $\bf W$. By taking $$ {\bf W}^T=\frac{\Sigma^{-1}{\bf\mu}^T}{{\bf1}\Sigma^{-1}{\bf\mu}^T}+{\bf \Delta}^T $$ we get $$ \left({\bf 1}\Sigma^{-1}{\bf\mu}^T\right)^2{\bf W}\Sigma\bf{W}^T ={\bf \Delta}\Sigma{\bf \Delta}^T\left({\bf 1}\Sigma^{-1}{\bf\mu}^T\right)^2 +2{\bf \Delta}{\bf\mu}\left({\bf 1}\Sigma^{-1}{\bf\mu}^T\right) +{\bf\mu}\Sigma^{-1}{\bf\mu}^T>0. $$ Recalling that $ax^2+bx+c>0$ implies $b^2-4ac<0$ we get the desired inequality.