No.
Inner products are bilinear maps, and a metric need not be. For example, the function
$d: V \times V \to \Bbb R$ given by:
$d(v,w) = 1$ if $v \neq w$
$d(v,v) = 0$
is a metric, but it most assuredly is *not* an inner product.
However...*given* an inner product $\langle \cdot,\cdot\rangle$, we can define a metric on an inner product space by:
$d(v,w) = \sqrt{\langle v-w,v-w\rangle}$
Metrics are a sort of "relaxing" of the requirements of the norm induced by an inner product-we can put a metric on an inner product space, but we may not be able to put an inner product on a metric space (such a space may not even have a vector addition defined on it).