The relation between trace and determinant of a matrix Let $M$ be a symmetric $n \times n$ matrix. 
Is there any equality or inequality that relates the trace and determinant of $M$?
 A: Not exactly what you're looking for but I would be remiss not to mention that for any complex square matrix $A$ the following identity holds:
$$\det(e^A)=e^{\mbox{tr}(A)} $$
A: The only relation I can think of is this one : Since the matrix $A$ is symetric, it is diagonalizable, thus it can be written $A = P^{-1}DP$ where $D$ is diagonal and $P$ is invertible. Therefore $$\det A = \prod_{i = 1} ^n \lambda_i$$ and $$tr A = \sum_{i = 1} ^n \lambda_i$$ where $\lambda_i$ is an eigenvalue.
Some inequalities can be found between the sum and the product.
A: The determinant and the trace are two quite different beasts, little relation can be found among them.
If the matrix is not only symmetric (hermitic) but also positive semi-definite, then its eigenvalues are real and non-negative.
Hence, given the properties ${\rm tr}(M)=\sum \lambda_i$ and ${\rm det}(M)=\prod \lambda_i$, and recalling the AM GM inequality, we get the following (probably not very useful) inequality:
$$\frac{{\rm tr}(M)}{n} \ge {\rm det}(M)^{1/n}$$
(equality holds iff $M = \lambda I$ for some $\lambda \ge  0$)
Much more interesting/insightful/useful are the answers by Owen Sizemore and Rodrigo de Azevedo.
A: The trace of $\bf M$ is the directional derivative of the determinant in the direction of $\bf M$ at ${\bf I}_n$, i.e.,
$$ \det \left( {\bf I}_n + h {\bf M} \right) = 1 + h \, \mbox{tr} ({\bf M}) + O \left( h^2 \right) $$
In Tao's words, "near the identity, the determinant behaves like the trace"$^\color{red}{\star}$. More generally,
$$\det( {\bf A} + h {\bf B} ) = \det({\bf A})  + h \, \mbox{tr} \left( \mbox{adj} ({\bf A}) \, {\bf B} \right) + O \left(h^2\right)$$
which is a variation of Jacobi's formula. Note that it is not required that $\bf M$ be symmetric.

$\color{red}{\star}$ Terence Tao, Matrix identities as derivatives of determinant identities, January 13, 2013.
A: No, there is not.  Consider the matrix with parameter $n$
$$\begin{bmatrix}
    1       & n \\
    n      &1 \\
\end{bmatrix}$$
The trace is 2, while the determinant is $1-n^2$.  You can vary $n$ to violate any possible inequality between the trace and the determinant.
A: Up to sign, the trace and determinant of an $n \times n$ matrix are coefficients of its characteristic polynomial (specifically, the coefficients in degrees $n-1$ and $0$ respectively).
The only constraint that the matrix being symmetric adds is that the characteristic polynomial is totally real — that is, all of its roots are real.
(note every totally real polynomial is a characteristic polynomial; e.g. of the diagonal matrix whose entries are the roots)
Thus, your problem is equivalent to

Let $f$ be a monic polynomial of degree $n$ whose roots are all real.  Is there any relationship between the coefficient in degree $n-1$ and the constant coefficient?

I believe you can only say anything when $n \leq 2$. When $n=2$, the requirement that the roots be real implies that $\mathrm{tr}(M)^2 \geq 4 \det(M)$.
A: Due to OP's fairly general formulation there's diverse bunch of answers by now. In addition to these, I'd like to mention some concrete relations expressing the determinant in terms of traces. They hold without the symmetry hypothesis, just assume dealing with a general complex matrix.
Despite being "quite different beasts", both $\det(M)$ and $\operatorname{tr}(M)$ of an $n\times n$ matrix $M$ are given by some $n$-variable symmetric polynomial, evaluated at the eigenvalues of $M$.

For any fixed matrix size $n$ there's a polynomial in $n$ variables, of total degree $n$, which yields $\det(M)$ when evaluated at
  $\left(\operatorname{tr}(M^n),\operatorname{tr}(M^{n-1}),\ldots,\operatorname{tr}(M)\right)$:
  $$
\det M\; =\;
\begin{cases}
\operatorname{tr}M & n=1\\[1.67ex]
\frac{1}{2}\big((\operatorname{tr}M)^2
 - \operatorname{tr}\left(M^2\right)\big) & n=2\\[1.67ex]
\frac{1}{6}\big((\operatorname{tr}M)^3
 - 3\operatorname{tr}\left(M^2\right)(\operatorname{tr}M)
 + 2\operatorname{tr}\left(M^3\right)\big) & n= 3\\[1.67ex]
 \quad\ldots & n\ge4
\end{cases}
$$

Case $\mathbf{n=1}$ is clear.
$\mathbf{n=2}$ is straightforward when applying Cayley-Hamilton to $M$
$$
M^2\:-\:(\operatorname{tr}M)\:M\:+\:(\det M)\pmatrix{1&0\\ 0&1}\;=\;\pmatrix{0&0\\0&0}
$$
and taking the trace.
Referring to Paul's answer/example featuring
$M=\left(\begin{smallmatrix}1&k\\ k&1\end{smallmatrix}\right)$,
thus $M^2=\left(\begin{smallmatrix}1+k^2&2k\\ 2k&1+k^2\end{smallmatrix}\right)$,
one obtains
$$\det M\;=\;\frac{1}{2}\big(4-2(1+k^2)\big) = 1-k^2$$
for the sake of illustration, presumably not the quickest way towards $\det(M)$
(the parameter has been renamed to $k$ since $n$ denotes the matrix size).
Cases $\mathbf{n\ge3}$ get more expensive $\ldots$ and are available:
A suitable entry point is the corresponding subsection in the Wikipedia entry on determinants.

If you clicked the preceding link then you may scroll down just a bit to get into a det-tr-inequality for a positive-definite matrix, also worthwhile as answer to the OP. Or skip directly down to it ;-) 
