Let $V$ be a Hilbert space, $\mathcal{A} : V \rightarrow V$ be a bounded linear operator satisfying that $(\mathcal{A}u,v) = (u,\mathcal{A}v)$, and $(\mathcal{A}u,u)>0$ if $u \neq 0$. I know that $\mathcal{A}$ is injective because $\mathcal{A}u = 0$ implies $u = 0$.
Does it hold that $\mathcal{A}$ is surjective? (Maybe use the fact that $\mathcal{A}$ is self-adjoint?)
It's easy to see that $\mathcal{A}$ induces another inner product $(u,v)_{\mathcal{A}} = (\mathcal{A}u,v)$. For any bounded linear operator $\mathcal{B}: V \rightarrow V$ with the original inner product, I am not sure the existence and the uniqueness of the adjoint operator: $$ (B^*u,v)_{\mathcal{A} }= (u, \mathcal{B}v)_{\mathcal{A} }. $$
I know the answer is yes when $V$ is finite-dimensional.
Consider the example where $V = l^2$ and $A : V \to V$ is the map $$ (x_1, x_2, x_3, \dots) \mapsto (x_1, \tfrac 1 2 x_2, \tfrac 1 3 x_3, \dots).$$
$A$ is a bounded linear operator, with norm $1$. $A$ is self-adjoint. And $\langle u, Au \rangle > 0$ for all non-zero $u \in V$.
But $v = (1, \tfrac 1 2, \tfrac 1 3, \dots)$ is an element of $V$ that is not in the image of $A$. Indeed, if $w$ is an element that is mapped by $A$ to $v$, then $w$ would have to be $(1, 1, 1, \dots)$. But $(1, 1, 1, \dots)$ is not actually an element of $l^2$, since it has an infinite $l^2$ norm.
Edit: Although your hypotheses do not imply that $A$ is surjective, it's worth knowing that they imply a weaker statement, which is that the image of $A$ is dense in $V$.
To prove this, consider a $v \in \text{Im}(A)^\perp$. For all $u \in V$, we have $\langle Au, v \rangle = 0$. Taking $u = v$, we see that $\langle Av , v \rangle = 0$. But one of your hypotheses states that the only $v$ that satisfies $\langle Av , v \rangle = 0$ is the zero vector. Thus we've proved that $\text{Im}(A)^\perp = \{ 0 \}$.
[Alternatively, you can prove the same thing using the weaker hypotheses that $A$ is injective and self-adjoint. If $v \in \text{Im}(A)^\perp$, then $\langle Au, v \rangle = 0$ for all $u \in V$. Since $A$ is self-adjoint, this is the same as saying that $\langle u, Av \rangle = 0$ for all $u \in V$. Taking $u = Av$, we obtain $\langle Av, Av \rangle = 0$, i.e. $Av = 0$. Since $A$ is injective, $v$ must be $0$. Thus $\text{Im}(A)^\perp = \{ 0 \}$.]
Anyway, if $W$ is any closed linear subspace of $V$, then $V$ can be written as a direct sum $V = W \oplus W^\perp$. Taking $W$ to be $\overline{\text{Im}(A)}$, we have $W^\perp = \overline{\text{Im}(A)}^\perp = \text{Im}(A)^\perp = \{ 0 \}$. So $V = \overline{\text{Im}(A)} \oplus \{ 0 \} = \overline{\text{Im}(A)}$. Thus $\text{Im}(A)$ is dense in $V$.
At this point you might like to return to our example where $V = l^2$ and where $A$ is the map $(x_1, x_2, x_3, \dots) \mapsto (x_1, \tfrac 1 2 x_2, \tfrac 1 3 x_3, \dots)$. Earlier, we showed that $v = (1, \tfrac 1 2, \tfrac 1 3, \dots)$ is not in $\text{Im}(A)$. However, $(1, \tfrac 1 2, \tfrac 1 3, \dots)$ is most certainly contained in the closure, $\overline{\text{Im}(A)}$; indeed, the elements $v_n := (1, \tfrac 1 2, \dots, \tfrac 1 n, 0, 0, \dots)$ are all in $\text{Im}(A)$, and the sequence $\{ v_n \}_{n \in \mathbb N}$ converges to $v$ in $l^2$.