positive operator is self-adjoint? Let $H$ be a Hilbert space, $A$ is a bounded operator on $H$ satisfies
\begin{equation}
(Ax,x) \ge 0, \; \forall x \in H,
\end{equation}
then how can we check that $A$ is symmetric?
 A: As pointed out by @ArcticChar, this is false over the field of real numbers.  It is however true over the complexes.  The proof is to write $\langle A(x),y\rangle$ using the complex polarization formula and then check that the resulting sesqui-linear form is Hermitian.
A: As pointed out by @Arctic Char and reiterated by @Ruy, this is false for real vector spaces. Consider, for example, $H=\mathbb{R}^2$ as a vector space over $\mathbb{R}$, and $A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. $A^T \neq A$, but $(Ax,x)\geq0$ (in fact, equal to $0$) for all $x$ in $H$.
Your statement holds true only for complex Hilbert spaces: that is, if $H$ is a complex Hilbert space, and $A$ is a bounded operator on $H$, if $(Ax,x)\geq 0$ for all $x$ in $H$, then $A$ is self-adjoint / hermitian.
Now to prove this: $(Ax,x)\geq 0$ implies that $(Ax,x)$ is real, that is, $(Ax,x)=\overline{(Ax,x)}=(x,Ax)=(A^\star x,x)$, and hence, $(Bx,x)=0$ for all $x$ in $H$, where $B=A-A^\star$. It can be seen (as in here) using the polarization identity that this implies $B=0$: for all $x,y$ in $H$,
\begin{equation}
(Bx,y)=\frac{1}{4}\left((B(x+y),x+y)-(B(x-y),x-y)+i(B(x+iy),x+iy)-i(B(x-iy),x-iy)\right)
\end{equation}
The right hand side is $0$, which implies that $B=0$, which in turn implies that $A=A^\star$.
