For two p.d. matrices $A$ and $B$, prove that $\lambda_1(AB)\leqslant \lambda_1(A) \cdot\lambda_1(B)$ If $A$ and $B$ are two nxn positive definite matrices, then show that

$$\lambda_1(AB) \leqslant \lambda_1(A) \cdot \lambda_1(B),$$

where $\lambda_1(\cdot)$ denotes the largest eigenvalue.
Let $C$ be a positive symmetric $n\times n$ matrix, and let $|\cdot|$ be the standard euclidean norm on $\Bbb R^n$. For all $X\in\Bbb R^n$,
$$|CX|\leq\lambda_1(C)|X|$$
with equality iff $X$ is eigenvector of $C$ with eigenvalue $\lambda_1(C)$, so that, again, for $C$ positive symmetric,
$$\lambda_1(C)=\max_{|X|\leq 1}|CX|=\|C\|$$ where $\|\cdot\|$ is the matrix norm subordinate to the standard euclidean norm on $\Bbb R^n$.
The answer to your question now follows, as for any matrix $D$ and any eigenvalue $\lambda$,  $|\lambda|\leq \|D\|$ so that
$$|\lambda_1(AB)|\leq\|AB\|\leq\|A\|\cdot\|B\|=\lambda_1(A)\lambda_1(B)\,.$$
Let the eigen values of $AB$ be $c_1, c_2, \ldots,c_n$ and those of $A$ and $B$ be $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ respectively.
Now, we know that, $\det(A)$ is the product of the eigenvalues of $A$, counted with their algebraic multiplicities.

Also, we have, $\det(AB) = \det(A)\det(B)$

$$
\Rightarrow c_1 c_2 \cdots c_n = (a_1 a_2 \cdots a_n)(b_1 b_2 \cdots b_n)
$$

=> $c_1$ * $c_2$ * ... * $c_n$ = ($a_1$ * $a_2$ * ... * $a_n$) * ($b_1$ * $b_2$ * ... * $b_n$ )

Using the above equality/identity(since all eigenvalues are positive, $A$ &amp; $B$ being p.d.), we can evidently write:

$$\lambda_1(AB) \le \lambda_1(A) \lambda_1(B).$$