Prove Inequality (with norm and an inner product) 
Prove that for all $f,g\in C[a,b]$ with the inner product $\langle f,g\rangle=\int_{a}^{b}f(x)\overline{g(x)}dx$:
$$\frac{1}{2}\int_{a}^{b}\int_{a}^{b}|f(x)g(y)-g(x)f(y)|^2dxdy\leq ||f||^2||g||^2-|\langle f,g\rangle^2|$$

I have started with $$\frac{1}{2}\int_{a}^{b}\int_{a}^{b}|f(x)g(y)-g(x)f(y)|^2dxdy=\langle f(x)g(y)-g(x)f(y),f(x)g(y)-g(x)f(y)\rangle\leq$$
By cauchy-schwarz
$$\leq ||f(x)g(y)-g(x)f(y)||\cdot||f(x)g(y)-g(x)f(y)||$$
Can we continue with
$$||f(x)g(y)-g(x)f(y)||\cdot||f(x)g(y)-g(x)f(y)||\leq [||f(x)g(y)||+||g(x)f(y)||]\cdot[||f(x)g(y)||+||g(x)f(y)||]?$$
Second thought: to try to prove
$$\frac{1}{2}\int_{a}^{b}\int_{a}^{b}|f(x)g(y)-g(x)f(y)|^2dxdy+|\langle f,g\rangle^2|\leq ||f||^2||g||^2$$
 A: It should be an equality:
\begin{align}
 &\frac{1}{2} \int \int \lvert f(x)g(y)-g(x)f(y) \rvert^2 \, dx \, dy \\
&= \frac{1}{2} \int \int \left( \lvert f(x) \rvert^2 \lvert g(y) \rvert^2 + \lvert g(x) \rvert^2 \lvert f(y) \rvert^2 - 2\Re(\overline{f(x)g(y)}g(x)f(y)) \right) dx \, dy \\
&= \lVert f \rVert^2 \lVert g \rVert^2 - \Re\left( \int \int \overline{f(x)g(y)}g(x)f(y) \, dx \, dy \right) \\
&= \lVert f \rVert^2 \lVert g \rVert^2 - \Re\left(\left( \int \overline{f(x)}g(x) \, dx \right) \left( \int \overline{g(y)}f(y)\, dy \right)\right) \\
&= \lVert f \rVert^2 \lVert g \rVert^2 - \lvert\langle f,g\rangle \rvert^2
\end{align}
(by expanding, linearity of the integral to bring out the real part, separating the $x$ and $y$ terms and using the definition of the inner-product).
A: In fact, equality holds:
$\newcommand\inner[2]{\left\langle #1, #2 \right\rangle}$
$$\frac{1}{2}\int_{a}^{b}\int_{a}^{b}|f(x)g(y)-g(x)f(y)|^2\,dxdy$$
$$ = \frac{1}{2}\int_{a}^{b}\int_{a}^{b}(f(x)g(y)-g(x)f(y))\overline{(f(x)g(y)-g(x)f(y))}\,dxdy$$
$$= \frac{1}{2}\int_{a}^{b}\int_{a}^{b}\left(|f(x)g(y)|^2 + |g(x)f(y)|^2 - f(x)\overline{g(x)}\overline{f(y)}g(y) - \overline{f(x)}g(x)f(y)\overline{g(y)}\right)\, dxdy$$
$$=\frac{1}{2}\int_{a}^{b} |g(y)|^2 \|f\|^2_2\,dy + \frac{1}{2}\int_{a}^{b} |f(y)|^2 \|g\|^2_2\,dy - \frac{1}{2}\int_{a}^{b} \overline{f(y)}g(y) \inner{f}{g}\,dy - \frac{1}{2}\int_{a}^{b} f(y)\overline{g(y)} \inner{g}{f}\,dy$$
$$= \|f\|^2_2\|g\|^2_2 - \inner{f}{g}\inner{g}{f}$$
$$= \|f\|^2_2\|g\|^2_2 - \left|\inner{f}{g}\right|^2$$
