<s>[INST] How prove this $\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\frac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$ 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  
 [/INST] 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. 
</s>