In my mathematical methods for physics course notes I find this:
A positive sesquilinear form is nondegenerate and Hermitian
The first statement is trivial: a positive form is such that
$\begin{cases}\hat{B}(x)\geqslant0\ \forall\,x\in X\\\hat{B}(x)=0\iff x=0\end{cases}$
Where $\hat{B}(x)=B(x,x)$ is the quadratic form associated to the sesquilinear form $B$.
The second condition gives us the nondegenerate part. If the form wasn't nondegenerate, there would be a $x\in X, x\ne0$ such that $B(x,y)=0\ \forall\,y\in X$. That means than even $B(x,x)=\hat{B}(x)=0$, which gives us an absurd with the fact $x$ must be $0$.
What I can't get is how the positivity (probably, the first condition of it) can cause the Hermitianity.
$\hat{B}(x)\geqslant0\ \forall\,x\in X\Rightarrow B(x,y)=\overline{B(y,x)}?$
Let $x,y \in X$, then we have \begin{align*} \sum_{k=0}^3 i^k B(x+i^k y, x+i^ky) &= B(x+y, x+y) - B(x-y,x-y) \\ & \quad + i \left[B(x+iy,x+iy) - B(x-iy, x-iy) \right] \\ &= 2 B(y,x)+2B(x,y) + 2B(x,y)-2B(x,y) + 2 B(y,x) \\ &= 4B(y,x). \end{align*} Since $ B(x+i^k y, x+i^ky) \in \mathbb{R}$, we have \begin{align*} 4\overline{B(y,x)} &=\sum_{k=0}^3 (-i)^k B(x+i^k y, x+i^ky) \\ &= \sum_{k=0}^3 i^k B(y+i^kx,y+i^kx)\\ &=B(x,y) \end{align*}