Actually i'm interested in the answer of this question and precisely i like to know how they achieved the following inequality:
$$\lambda_k(A)\lambda_1(B)\leq \lambda_k(AB)$$
based on this trick:
\begin{align*}
\lambda_k(AB)=\lambda_k(\sqrt{B}A\sqrt{B})
&=\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(\sqrt{B}A\sqrt{B}x,x)}{(x,x)}\right)\\
&=\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}
\frac{(Bx,x)}{(x,x)}\right).
\end{align*}
Best Answer
So, the assumption is that $A,B$ are symmetric positive-definite. Then each has asymmetric positive-definite square root. Since the eigenvalues of the product do not depend on order, $$\lambda_k(AB)=\lambda_k(\sqrt B\,A\,\sqrt B).$$ The next two equalities are the min-max theorem.
We have, since $B$ is symmetric, $$ \lambda_1(B)\leq\frac{\langle Bx,x\rangle}{\langle x,x\rangle}\leq\lambda_n(B) $$ for all $x\ne0$.
So $$ \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)} \frac{(Bx,x)}{(x,x)} \geq\lambda_1(B)\, \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)} $$ Now as you take the minimum over all $F$ of dimension $k$, since $\sqrt B$ is invertible you get \begin{align} \min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)} \frac{(Bx,x)}{(x,x)}\right) &\geq\lambda_1(B)\,\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}\right)\\ \ \\ &=\lambda_1(B)\,\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(Ax,x)}{(x,x)} \right)\\ \ \\ &=\lambda_1(B)\lambda_k(A). \end{align}