Let $A,B$ be $n \times n$ Hermitian matrices whose eigen values are non-negatives. Let $\lambda_1(X) \geq
\lambda_1(X) \geq \lambda_2(X) \geq \cdots \geq \lambda_n(X)$ denote
ordered eigenvalues of $n \times n$ Hermitian matrix $X$.

Then, $\lambda_i(AB) \leq \lambda_i(A)\lambda_{j-i+1}(B),$
for any $i \leq j$.

Proof.
Let $a_i,b_i$ and $c_i$ be the unitary eigenvectors of $A,B$ and $A+B$, respectively. Namely, $A a_i = \lambda_i(A)a_i$ and so on.

Let $V_a,V_b$ and $V_c$ are vector spaces spanned by the vectors $\{a_i,...,a_n\}$ and $\{b_{j-i+1},...,b_n\}$ and $\{c_1,...,c_j\}$, respectively. Then, $V_a \cap V_b \cap V_c \neq 0$. In fact, for arbitrary vector spaces $X,Y$, we can say $\dim (X \cap Y) = \dim(X)+\dim(X) -\dim(X+Y )$ and using this twice, it follows that

$\dim(V_a \cap V_b \cap V_c ) =\dim(V_a \cap( V_b \cap V_c) )$

$=\dim(V_a) + \dim( V_b \cap V_c) - \dim(V_a + ( V_b \cap V_c)) $

$=\dim(V_a) + \dim( V_b ) + \dim( V_c) - \dim( V_b +V_c) - \dim(V_a + ( V_b \cap V_c)) $

$\geq\dim(V_a) + \dim( V_b ) + \dim( V_c) - n -n $

$= (n-i+1) + (n-j+i-1)+1 + j - n -n$

$= 1,$

and thus there is a nonzero unit vector $x \in V_a \cap V_b \cap V_c.$

Because $x \in V_c$, we can write $x= \sum_{\nu =1}^j x^\nu c_\nu$ and then,

$<x,(AB)x>= < \sum_{\nu =1}^j x^\nu c_\nu , (AB) \sum_{\nu =1}^j x^\nu c_\nu >$

$ = < \sum_{\nu =1}^j x^\nu c_\nu , \sum_{\nu =1}^j x^\nu \lambda_\nu(AB)c_\nu >$

$ = \sum_{\nu =1}^j | x^\nu |^2 \lambda_\nu(AB)$

$ \geq \sum_{\nu =1}^j | x^\nu |^2 \lambda_j(AB)$

$ = \lambda_j(AB) \cdots \cdots \cdots (1).$

Similarly, because $x \in V_a$, we can write $x= \sum_{\nu =i}^n x^\nu a_\nu$ and then,

$<x,Ax>= < \sum_{\nu =i}^n x^\nu a_\nu , A\sum_{\nu =i}^n x^\nu a_\nu >$

$ = < \sum_{\nu =i}^n x^\nu a_\nu , A\sum_{\nu =i}^n x^\nu a_\nu >$

$ = \sum_{\nu =i}^n | x^\nu |^2 \lambda_\nu(A )$

$ \leq \sum_{\nu =i}^j | x^\nu |^2 \lambda_i(A )$

$ = \lambda_i(A ) \cdots \cdots \cdots (2).$

Similarly,
$<x,Bx> \leq \lambda_{j-i+1}(B ) \cdots \cdots \cdots (3).$

Combining the inequalities (1),(2),(3) and Cauchy Schwartz inequality, we get

$\lambda_j(AB) \leq <x,(AB)x> \leq |<A^*x , Bx>| \leq |A^*x| \dot | Bx| \leq \lambda_i(A ) \lambda_{j-i+1}(B ). $

This completes the proof.