A property about positive definite matrix For any two positive definite same-sized matrices $A$ and $B$, I guessed a property about them and I have verified it via numerical experiment in matlab. The property is stated below:
Assume $A$ and $B$ are both $n\times n$ positive definite matrices, then we define a vector $\vec{V}$.
$$V(i)=A_{ii}B_{ii}$$
that is, the $i$th component of $\vec{V}$ is $A_{ii}B_{ii}$. The minimal component of $\vec{V}$ is:
$$V_{min}=min\{V_{i}:i=1,2,3...,n\}$$
Then we can find a bound for the eigenvalue of their product $AB$, denote the smallest eigenvalue of $AB$ $\lambda_{min}$.
The claim is that:
$$V_{min}\geq\lambda_{min}$$
Can someone prove it?

We have the following

Theorem. When $A,B$ are positive definite, $\lambda_\min(AB)\le\lambda_\min(A\circ B)$.

This is a nontrivial result. For a proof, see theorem 3b of C.R. Johnson and L. Elsner (1987), The Relationship between Hadamard and Conventional Multiplication for Positive Definite Matrices, Linear Algebra and Its Applications, 92:231-240.
As the Hadamard product $A\circ B$ is real symmetric, its minimum eigenvalue $\lambda_\min(A\circ B)$ is given by $\min_{\|v\|_2=1}v^T(A\circ B)v$. Hence it is bounded above by all diagonal elements of $A\circ B$.
