Inequality on norm in terms of eigenvalue I came across this inequality but not sure if it's true:
$|\lambda_{\text{min}}|^2\|\hat{\beta}-\beta\|^2_2\leq (\hat{\beta}-\beta)'\hat{\Sigma}(\hat{\beta}-\beta)$
where $\lambda_{\text{min}}$ is the min eigenvalue of $\hat{\Sigma}\equiv X'X/n$ for some $n\times k$ matrix $X$, and $\hat{\beta},\beta$ are $k\times 1$.
I take this inequality is equivalent to
$|\lambda_{\text{min}}|\|\hat{\beta}-\beta\|_2\leq \|X(\hat{\beta}-\beta)\|_2/n$, but not sure where to go from here.
How can the result be shown (if it is indeed true)? Or is something similar true?
Hard to give more context here---I came across it when studying class notes of compatibility constraint in LASSO; but hopefully it doesn't need the context to address.

I just want to add that there is also a connection to the Rayleigh quotient, which for the real symmetric $k \times k$ Matrix $\hat{\Sigma}$ is defined as $  \frac{x^\prime \hat{\Sigma} x }{x^\prime  x}, \,x \in \mathbb{R}^{k},\, x \neq 0$ .
By the Courant–Fischer Theorem, we obtain $$ \lambda_{min}  = \min_{x \neq 0} \frac{x^\prime \hat{\Sigma} x }{x^\prime  x}, \,x \in \mathbb{R}^{k},\, x \neq 0.$$
Let $\delta:= \hat{\beta}-\beta \in \mathbb{R}^{k}$ and $\delta\neq 0$. Then, \begin{equation} 
 \lambda_{min}  = \min_{x \neq 0} \frac{x^\prime \hat{\Sigma} x }{x^\prime  x}  \leq \frac{\delta^\prime \hat{\Sigma} \delta }{ \delta^\prime  \delta} = \frac{(\hat{\beta}-\beta )^\prime \hat{\Sigma} (\hat{\beta}-\beta ) }{ (\hat{\beta}-\beta )^\prime  (\hat{\beta}-\beta )}, \end{equation}
where the inequality holds since  $\delta  \in \mathbb{R}^{k}, \, \delta \neq 0,$ and we are minimizing over $x  \in \mathbb{R}^{k}, \, x \neq 0.$
Rearranging the inequality, we get
$$\lambda_{min} \, \Vert\hat{\beta}-\beta \Vert_2^2 \leq (\hat{\beta}-\beta )^\prime \hat{\Sigma} (\hat{\beta}-\beta ) .$$
(For $\delta=0$, the last inequality obviously also holds.)
The square in your inequality may be related to singular values? Let $\sigma_{min}$ denote the minimal singular value value of $ X/\sqrt{n}$, then $\lambda_{min} = \sigma_{min}^2$ and therefore,
$$\sigma_{min}^2  \, \Vert\hat{\beta}-\beta \Vert_2^2 \leq (\hat{\beta}-\beta )^\prime \hat{\Sigma} (\hat{\beta}-\beta ).$$
