I think this is false. Let $A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle < 0$.
The point is that the matrix can have all its eigenvalues strictly positive, but it does not follow that it is positive definite.
Let $A \in M_{n}(\mathbb{R})$ be any non-symmetric $n\times n$ matrix but "positive definite" in the sense that:
$$\forall x \in \mathbb{R}^n, x \ne 0 \implies x^T A x > 0$$
The eigenvalues of $A$ need not be positive. For an example, the matrix in David's comment:
$$\begin{pmatrix}1&1\\-1&1\end{pmatrix}$$
has eigenvalue $1 \pm i$. However, the real part of any eigenvalue $\lambda$ of $A$ is always positive.
Let $\lambda = \mu + i\nu\in\mathbb C $ where $\mu, \nu \in \mathbb{R}$ be an eigenvalue of $A$. Let $z \in \mathbb{C}^n$ be a right eigenvector associated with $\lambda$. Decompose $z$ as $x + iy$ where $x, y \in \mathbb{R}^n$.
$$(A - \lambda) z = 0 \implies \left((A - \mu) - i\nu\right)(x + iy) = 0
\implies \begin{cases}(A-\mu) x + \nu y = 0\\(A - \mu) y - \nu x = 0\end{cases}$$
This implies
$$x^T(A-\mu)x + y^T(A-\mu)y = \nu (y^T x - x^T y) = 0$$
and hence
$$\mu = \frac{x^TA x + y^TAy}{x^Tx + y^Ty} > 0$$
In particular, this means any real eigenvalue $\lambda$ of $A$ is positive.
Best Answer
Suppose our matrix $A$ has eigenvalue $\lambda$.
If $\lambda = 0$, then there is some eigenvector $x$ so that $Ax = 0$. But then $x^T A x = 0$, and so $A$ is not positive definite.
If $\lambda < 0$, then there is some eigenvector $x$ so that $Ax = \lambda x$. But then $x^T A x = \lambda \lvert x \rvert^2$, which is negative since $\lvert x \rvert^2 > 0$ and $\lambda < 0$. Thus $A$ is not positive definite.
And so if $A$ is positive definite, it only has positive eigenvalues.