An inequality on trace of product of two matrices Suppose we have two $n \times n$ positive semidefinite matrices, $A$ and $B$, such that $\mbox{tr}(A), \mbox{tr}(B) \le 1$. 
Can we say anything about $\mbox{tr}(AB)$? Is $\mbox{tr}(AB) \le 1 $ too?
 A: An extension to @dineshdileep 's answer can also be found here, in which it shows that:
$$tr(A^TB)\le \sqrt{tr(A^TA)tr(B^TB)}$$
A: In the space of positive semi-definite matrices, trace is a proper inner-product (it is easy to show that), i.e. it obeys the Cauchy-Schwarz inequality: $\langle x,y \rangle \leq \sqrt{ \langle x,x \rangle \langle y,y\rangle}$. So
$$\mbox{tr}\{AB\}\leq \sqrt{\mbox{tr}\{A^2\} \mbox{tr}\{B^2\}}$$
Now, since $A$ is positive semidefinite, $\mbox{tr}\{A^2\} \leq \mbox{tr}\{A\}^2$, i.e., the eigenvalues of $A^2$ are squared eigenvalues of $A$, and since they are positive
$$\mbox{tr} \{A^2\} = \sum_{i=1}^{N}\lambda_{i}^{2}\leq \left( \sum_{i=1}^{N}\lambda_{i} \right)^{2} = \mbox{tr}\{A\}^2 \leq 1$$ 
A similar argument for B proves $\mbox{tr}\{B^2\}\leq 1$ . So $\mbox{tr}\{AB\}\leq 1$. Hope this answers your question. 
A: Let's write $A = \sum_{i=1}^{n}\lambda_i u_iu_i^T$ and $B = \sum_{i=1}^{n}\mu_i v_iv_i^T$.
$$AB = \sum_{i,j} \lambda_i \mu_j u_iu_i^Tv_jv_j^T$$
and so $$\text{Tr}(AB) = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j \left|u_i^Tv_j\right|^2 \le \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j \left\|u_i\right\|^2\left\|v_j\right\|^2 = \sum_{i=1}^n \lambda_i \sum_{j=1}^n\mu_j = \text{Tr}(A)\text{Tr}(B)$$
A: First, note that $(A-B)^{2}$ is positive semi-definite, so we have:
$$0\leq\mathrm{Tr}(A-B)^{2}=\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2})-2\mathrm{Tr}(AB)$$
$$\mathrm{Tr}(AB)\leq\frac{1}{2}(\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2}))$$
Second, for $A$ positive semi-definite, suppose that all of eigenvalues are $\lambda_{1}$, $\lambda_{2}$, $\cdots$, $\lambda_{n}$, then $\lambda_{i}\geq0$ and $\mathrm{Tr}A=\sum_{i=1}^{n}\lambda_{i}\leq1$, so $\mathrm{Tr}(A^{2})=\sum_{i=1}^{n}\lambda_{i}^{2}\leq\sum_{i=1}^{n}\lambda_{i}\leq1$.
Similarly, $\mathrm{Tr}(B^{2})\leq1$, so $\mathrm{Tr}(AB)\leq1$.
Remark. More generally, we can conclude that the range of $\mathrm{Tr}(AB)$ is $[0,1]$.
As $A$ and $B$ are positive semi-definite, so there exist $C$ and $D$ such that $A=C^{T}C$ and $B=D^{T}D$, so $\mathrm{Tr}(AB)=\mathrm{Tr}(C^{T}CD^{T}D)=\mathrm{Tr}(CD^{T}DC^{T})=\mathrm{Tr}[CD^{T}(CD^{T})^{T}]\geq0$.
Set $A=diag[1,0,0,\cdots,0]$ and $A=diag[0,1,0,\cdots,0]$, then $\mathrm{Tr}(AB)=0$.
Set $A=B=diag[1,0,0,\cdots,0]$, then $\mathrm{Tr}(AB)=1$.
Accoading to above, we can conclude that $\text{Range}(\mathrm{Tr}(AB))=[0,1]$.
A: When $A$ and $B$ are positive semidefinite, so are $\operatorname{tr}(A)I-A$ (by considering its eigenvalues) and $B^{1/2}\left(\operatorname{tr}(A)I-A\right)B^{1/2}$. Hence
$$
\operatorname{tr}(A)\operatorname{tr}(B)-\operatorname{tr}(AB)
=\operatorname{tr}\big(\left(\operatorname{tr}(A)I-A\right)B\big)
=\operatorname{tr}\left(B^{1/2}\left(\operatorname{tr}(A)I-A\right)B^{1/2}\right)
\ge0
$$
and the result follows.
Using the same idea, one can actually obtain a tighter inequality $\operatorname{tr}(AB)\le\rho(A)\operatorname{tr}(B)$.
