$\omega$-area of a unit circle - Symplectic geometry Consider a standard symplectic structure $\omega=\sum_k dp_k\wedge dq_k$ on $\mathbb{R}^{2n}$. Let $S$ be a unit circle that belongs to some plane in $\mathbb{R}^{2n}$. I want to show that
$$\iint_S \omega\leq \pi.$$
One of the hint was to use Cauchy-Shwartz inequality to corresponding Hermitian form
$$\langle z,w\rangle=\sum z_k\overline{w_k}.$$
But, unfortunately, I don't see how to proceed further. Any hint would be appreciated. Are there any other approach to this problem?
Also, as I understand it geometrically, we basically compute the area of projection of the circle onto $p_iq_i$-planes, correct?

Using the complex coordinates $z_i = p_i + \sqrt{-1} q_i$ and its conjugate $\bar z_i = p_i - \sqrt{-1} q_i$, the symplectic form can be written as
$$ \omega = \frac{\sqrt{-1}}{2}\sum_i dz_i \wedge d\bar z_i .$$
Thus for any $u = (u_1, \cdots, u_n)$ and $v = (v_1, \cdots, v_n)$ in $\mathbb C^n$,
\begin{align} 
\omega (u, v) &= \frac{\sqrt{-1}}{2} \sum_i (u_i \bar v_i - \bar u_i v_i)\\
&= - \operatorname{Im} \langle u, v\rangle 
\end{align}
Now let $e_1, e_2$ be two orthonormal basis in the plane which contains the unit disc. Then
$$ \iint_D \omega = \iint_D \omega (e_1, e_2) dxdy \le \iint _D |\langle e_1 , e_2\rangle| dxdy\le \iint _D \|e_1\| \| e_2\| dxdy =\pi$$
