Prove that there is no function $f$ such that $\int_{0}^{\pi} |f(x)-\sin x|^2dx \le \frac34$ and $\int_{0}^{\pi} |f(x)-\cos x|^2 dx \le \frac34$ 
Prove that there does not exist a continuous function $f:[0,\pi] \to \mathbb{R}$ such that $$\int_{0}^{\pi} |f(x)-\sin x|^2 \mathrm{d}x \le \frac34  \\ \text{and} \\ \int_{0}^{\pi} |f(x)-\cos x|^2 \mathrm{d}x \le \frac34$$

I thought of some inequalities like cauchy-schwarz and AM-GM , but it didn't help. Also, since there is no condition on $f(x)$ , it could be positive or negative, so that makes it a bit challenging.
Thank you for helping!
 A: Assume such a function exists. We have
$$
|\cos x - \sin x| \le |f(x)-\cos x| + |f(x) - \sin x|
$$
by the triangle inequality. Hence,
$$
\begin{align*}
|\cos x - \sin x|^2 \le
|f(x)-\cos x| ^2 + |f(x) - \sin x|^2 + 2|f(x)-\cos x||f(x) - \sin x| 
\end{align*}
$$
Integrating from $0$ to $\pi$ and applying the Cauchy-Schwarz inequality we get $\pi \le 3$, a contradiction.
A: Consider $L_2$ metric $d(f,g)$ on $(0,\pi)$.  $\int\limits_0^{\pi}(\sin(x)-\cos(x))^2dx=\pi$ or $d(\cos(x),\sin(x))=\sqrt{\pi}$.  However $d(\cos(x),\sin(x))\le d(\cos(x),f(x))+d(\sin(x),f(x))$.  If both integrals are $\le \frac{3}{4}$, then $2\sqrt{\frac{3}{4}}\ge \sqrt{\pi}$ or $1.732 \ge 1.772$, contradiction.
A: Alternative approach: assume that
$$ f(x)\stackrel{L^2(0,\pi)}{=} c_0 + \sum_{n\geq 1} s_n \sin(2nx) + \sum_{n\geq 1} c_n\cos(2nx). $$
By elementary integrals we have
$$ \sin(x) = \frac{2}{\pi}-\sum_{n\geq 1}\frac{4}{\pi(4n^2-1)}\cos(2nx) $$
$$ \cos(x) = \sum_{n\geq 1}\frac{8n}{\pi(4n^2-1)}\sin(2nx) $$
so
$$ f(x)-\sin(x) = \left(c_0-\frac{2}{\pi}\right)+\sum_{n\geq  1}s_n \sin(2nx)+\sum_{n\geq 1}\left(c_n+\frac{4}{\pi(4n^2-1)}\right)\cos(2nx) $$
$$ f(x)-\cos(x) = c_0 + \sum_{n\geq 1}\left(s_n-\frac{8n}{\pi(4n^2-1)}\right)\sin(2nx)+\sum_{n\geq 1}c_n \cos(2nx)$$
and the given constraints can be represented as
$$ \pi\left(c_0-\frac{2}{\pi}\right)^2 + \frac{\pi}{2}\sum_{n\geq 1}s_n^2 + \frac{\pi}{2}\sum_{n\geq 1}\left(c_n+\frac{4}{\pi(4n^2-1)}\right)^2 \leq \frac{3}{4} $$
$$ \pi c_0^2 +\frac{\pi}{2}\sum_{n\geq 1}\left(s_n-\frac{8n}{\pi(4n^2-1)}\right)^2 + \frac{\pi}{2}\sum_{n\geq 1}c_n^2 \leq \frac{3}{4} $$
leading to
$$ 2\left[c_0^2+\left(c_0-\frac{2}{\pi}\right)^2\right]+\sum_{n\geq 1}\left[s_n^2+\left(s_n-\frac{8n}{\pi(4n^2-1)}\right)^2 \right]+\sum_{n\geq 1}\left[c_n^2+\left(c_n+\frac{4}{\pi(4n^2-1)}\right)^2\right] \leq \frac{3}{2\pi}$$
The LHS is a quadratic form in $c_0,c_1,\ldots,s_1,s_2,\ldots$ and a convex function with respect to any of its variables. It follows that the LHS has a unique minimum, attained at $c_0=\frac{1}{\pi},s_n=\frac{4n}{\pi(4n^2-1)}$ and $c_n=-\frac{2}{\pi(4n^2-1)}$. In equivalent terms
$$ \text{argmin}\int_{0}^{\pi}\left[(f(x)-\sin(x))^2+(f(x)-\cos(x))^2\right]\,dx = \frac{\sin(x)+\cos(x)}{2} $$
but in such a case
$$ \int_{0}^{\pi}(f(x)-\sin(x))^2\,dx = \int_{0}^{\pi}(f(x)-\cos(x))^2\,dx = \frac{\pi}{4} > \frac{3}{4}. $$
