Eigenvalue inequality $\lambda_{\min}(AB) \geq \lambda_{\min}(A)\lambda_{\min}(B)$ 

$A, B$ are $n\times n$ positive semidefinite Hermite matrices.  $\lambda_{\min}(A)$ denotes the minimum eigenvalue of $A$. then we have 

$$\lambda_{\min}(AB) \geq \lambda_{\min}(A)\lambda_{\min}(B)$$


en, $\lambda_{\min}(A)=\min\limits_{x\ne0}\dfrac{x^*Ax}{x^*x}$. But how to go on?
Thanks a lot!
 A: Hint. Suppose you can show that $\lambda_\min(M) = \min_{x\ne0} \frac{\|Mx\|}{\|x\|}$ whenever $M$ is positive definite. It follows that $\lambda_\min(M)\|x\| \le \|Mx\|$ for every vector $x$. Therefore,
$$
\lambda_\min(A)\lambda_\min(B)\|x\|
\le \lambda_\min(A)\|Bx\|
\le \|ABx\|
$$
for every nonzero vector $x$. Take $x$ as a unit eigenvector corresponding to the minimum eigenvalue (whatever it means) of $AB$, we get $\lambda_\min(A)\lambda_\min(B)\le |\lambda_\min(AB)|$. Note that for any two complex square matrices $X$ and $Y$ of the same sizes, $XY$ and $YX$ have identical characteristic polynomials and identical spectra. Hence $\lambda_\min(AB)=\lambda_\min(A^{1/2}BA^{1/2})$. Can you show that $\lambda_\min(A^{1/2}BA^{1/2})\ge0$?
A: (1) First, let us prove the inequality $\lambda_{min}(DB)\geq\lambda_{min}(D)\lambda_{min}(B)$, when $D$ is a positive definite and $B$ is a positive semidefinite Hermitian matrices.
Let us write $D=C^2$, where $C$ is also a positive definite Hermitian matrix.
Thus, $det(DB-\lambda Id)=det(C^2B-\lambda Id)=det(C^{-1}(C^2B-\lambda Id)C)=det(CBC-\lambda Id)$.
Therefore, the characteristic polynomial of $DB$ is equal to the characteristic polynomial of $CBC$, which is hermtian and positive semidefinite.
The minimal eigenvalue of $DB$ is the minimal eigenvalue of $CBC$, which is $\min_{x\neq 0} \dfrac{x^*CBCx}{x^*x}$.
Now, for every $x\neq 0$,
$$\dfrac{x^*CBCx}{x^*x}=\dfrac{x^*CBCx}{x^*CCx}\dfrac{x^*CCx}{x^*x}\geq (\min_{z\neq 0} \dfrac{z^*CBCz}{z^*CCz})(\min_{w\neq 0} \dfrac{w^*CCw}{w^*w})=(\min_{y\neq 0} \dfrac{y^*By}{y^*y})(\min_{w\neq 0} \dfrac{w^*Dw}{w^*w}).$$
Therefore, $$\lambda_{min}(DB)=\lambda_{min}(CBC)=\min_{x\neq 0}\dfrac{x^*CBCx}{x^*x}\geq (\min_{y\neq 0} \dfrac{y^*By}{y^*y})(\min_{x\neq 0} \dfrac{w^*Dw}{w^*w})=\lambda_{min}(B)\lambda_{min}(D).$$
(2) Now, let us prove that if A is positive semidefinite (with an eingenvalue equal to zero) and $B$ is positive semidefinite then $\lambda_{min}(AB)=0=\lambda_{min}(A)\lambda_{min}(B)$.
Let $A=U\left(\begin{array}{cc}
D_{k\times k} & 0_{k\times s} \\ 
0_{s\times k} & 0_{s\times s}
\end{array} \right)U^*$, where $D$ is a positive definite real diagonal matrix and $U$ is unitary.
Now $AB$ have the same eigenvalues of $\left(\begin{array}{cc}
D_{k\times k} & 0_{k\times s} \\ 
0_{s\times k} & 0_{s\times s}
\end{array} \right)U^*BU$, because they are similar. 
Now $\left(\begin{array}{cc}
D & 0_{k\times s} \\ 
0_{s\times k} & 0_{s\times s}
\end{array} \right)U^*BU=\left(\begin{array}{cc}
DB_1 & DB_2 \\ 
0_{s\times k} & 0_{s\times s}
\end{array} \right)$, where $B_1$ is a positive semidefinite Hermitian matrix as a principal submatrix of $U^*BU$.
Now, the eigenvalues of  $\left(\begin{array}{cc}
DB_1 & DB_2 \\ 
0_{s\times k} & 0_{s\times s}
\end{array} \right)$ are the eigenvalues of  $DB_1$ and zero. Therefore the eigenvalues of $AB$ are the eigenvalues of  $DB_1$ and zero.
But $\lambda_{min}(DB_1)\geq\lambda_{min}(D)\lambda_{min}(B_1)\geq 0$, by item (1).
Thus, the minimal eigenvalue of $AB$ is zero and $\lambda_{min}(A)\lambda_{min}(B)=0$.
