Question: Let $M\_1$ and $M\_2$ be two symmetric $d\times d$ matrices. What is the relationship between
$tr(M\_1M\_2M\_1M\_2)$ and $tr(M\_1^2 M\_2^2 )$?

P.S. I tried a few examples and found
$$
tr(M\_1M\_2M\_1M\_2) \le tr(M\_1^2 M\_2^2 )
$$
seems always true. Is there a theorem?

Answer: 
Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality.

Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we have 
\begin{equation\*}
 |\text{tr}(AB)^{2p}| \le \text{tr}A^{2p}B^{2p}
\end{equation\*}
Not to take anything away from Suvrit's answer, but this is actually much simpler. First, we can assume $M\_1$ is diagonal. Call it $diag(x\_1, \dotsc, x\_i).$
Then the difference between the LHS and the RHS is 

$$\sum\_{i> j} a\_{ij}^2 (x\_i - x\_j)^2,$$
where the $a\_{ij}$ are the entries of $M\_2.$