Let $X$ be an N-dimensional random variable with mean $\mu=\mathbb{E}[X]$ and covariance matrix $V=\mathbb{E}\left[ \left( X-\mu\right) \, \left( X-\mu\right)^T\right]$

If $V$ is invertible (i.e., strictly positive), for any $t>0$ $$ \Pr\left( \sqrt{\left( X-\mu\right)^T \, V^{-1} \, \left( X-\mu\right) } > t \right) \le \frac{N}{t^2} $$

Proof: $V$ is positive, so $V^{-1}$ is. Define the random variable $$ y = \left( X-\mu\right)^T \, V^{-1} \, \left( X-\mu\right) $$ $y$ is positive, then Markov's inequality holds: $$ \Pr\left( \sqrt{\left( X-\mu\right)^T \, V^{-1} \, \left( X-\mu\right) } > t\right) = \Pr\left( \sqrt{y} > t\right) =\Pr\left( y > t^2 \right) \le \frac{\mathbb{E}[y]}{t^2} $$

Since $V$ is symmetric, a rotation $R$ (i.e., $R\, R^T = R^T\, R = I$ and a diagonal matrix $D$ (i.e., $i\neq j \, \Rightarrow \, D_{i,j}=0$ exist such that $$ V = R^T \, D \, R $$ Since $V$ is positive $D_{ii}>0$ Besides $$ V^{-1} = R^{-1} \, D^{-1} \, (R^T)^{-1} = R^T \, D^{-1} \, R $$ clearly $\left[ D^{-1}\right]_{ii} = \frac{1}{D_{ii}}$

Define $Z = R \, \left( X-\mu\right)$

The following identities hold: $$ \mathbb{E}\left[ Z \, Z^T \right] = R \,\mathbb{E}\left[ \left( X-\mu\right) \, \left( X-\mu\right)^T \right] \, R^T = R \, R^T \, D \, R \, R^T = D \quad \Rightarrow \quad \forall i \quad \mathbb{E}\left[ Z_i^2 \right] = D_{ii} $$ and $$ y = Z^T \, R \, V^{-1} \, R^T \, Z = Z^T \, D^{-1} \, Z = \sum\limits_{i=1}^N \frac{Z_i^2}{D_{ii}} $$ then $$ \mathbb{E} [y] = \sum\limits_{i=1}^N \frac{\mathbb{E}\left[ Z_i^2 \right] }{D_{ii}} = N $$