Prove or disprove:
if $f(x)\ge 0,\forall x\in [-\pi,\pi]$,show that
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\dfrac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$$
I can prove this if $2\pi$ takes the place of $\dfrac{\pi}{2}$
because use Cauchy-schwarz inequality we have
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2\le\int_{-\pi}^{\pi}\sin^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$
$$\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\int_{-\pi}^{\pi}\cos^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$
add this two inequality, we have
$$\begin{align*}\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2 &\le \int_{-\pi}^{\pi}f^2(x)dx\int_{-\pi}^{\pi}(\sin^2{x}+\cos^2{x})dx\\
&=2\pi\int_{-\pi}^{+\pi}f^2(x)dx\end{align*}$$
see this Discrete form of inequality：Prove this inequality with Cauchy-Schwarz inequality
So far, I haven't found any counterexamples，such $f(x)=1,\sin{x}+1$ it such this inequality
Best Answer
Note that \begin{align*} \int_{-\pi}^\pi f(x) \sin x \,dx &= \int_0^{\pi/2} (f(x) - f(x - \pi) + f(\pi - x) - f(-x)) \sin x \,dx \\ &= \int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx \end{align*} and similarly \begin{align*} \int_{-\pi}^\pi f(x) \cos x \,dx &= \int_0^{\pi/2} (f(x) - f(x - \pi) - f(\pi - x) + f(-x)) \cos x \,dx \\ &= \int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx \end{align*} where we define $g, h : [0, \pi/2] \to \mathbb{R}$ by $g(x) = f(x) - f(x - \pi)$ and $h(x) = f(\pi - x) - f(-x)$.
Then by Cauchy-Schwarz, \begin{align*} \left(\int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx\right)^2 &\leq \int_0^{\pi/2} \sin^2 x \,dx \int_0^{\pi/2} (g(x) + h(x))^2 \,dx \\ &= \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx \end{align*} and \begin{align*} \left(\int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx\right)^2 &\leq \int_0^{\pi/2} \cos^2 x \,dx \int_0^{\pi/2} (g(x) - h(x))^2 \,dx \\ &= \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx \end{align*} hence \begin{align*} \left(\int_{-\pi}^\pi f(x) \sin x \,dx\right)^2 &+ \left(\int_{-\pi}^\pi f(x) \cos x \,dx\right)^2 \\ &\leq \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx + \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx\\ &= \frac{\pi}{2} \int_0^{\pi/2} (g(x))^2 + (h(x))^2 \,dx \\ &\leq \frac{\pi}{2} \int_0^{\pi/2} (f(x))^2 + (f(x - \pi))^2 + (f(\pi - x))^2 + (f(-x))^2 \,dx \\ &= \frac{\pi}{2} \int_{-\pi}^\pi (f(x))^2 \,dx \end{align*} as desired, where the last inequality holds because $f$ is nonnegative.