Prove the upper bound of the minimum involving trigonometric functions Does anyone know if
$$\min\left(\begin{array}{c}
|\sin x|, |\sin y|, |\sin z|, |\cos x|, |\cos y|, |\cos z|, \\
|\sin(x-y)|, |\sin(x-z)|, |\sin(y-z)|,\\
|\cos(x-y)|, |\cos(x-z)|, |\cos(y-z)| 
\end{array}\right) \leq \frac25$$
for all real $x$, $y$, $z$?
Thanks a lot!
 A: Your claim is that for every $x,y,z \in \mathbb R$, at least one of the following inequalities is true:
\begin{align}\newcommand{ab}[1]{\lvert{#1}\rvert}
\ab{\sin x} &\leq \tfrac25, & 
\ab{\sin y} &\leq \tfrac25, & 
\ab{\sin z} &\leq \tfrac25, \\
\ab{\cos x} &\leq \tfrac25, & 
\ab{\cos y} &\leq \tfrac25, & 
\ab{\cos z} &\leq \tfrac25, \\
\ab{\sin(x - y)} &\leq \tfrac25, & 
\ab{\sin(x - z)} &\leq \tfrac25, & 
\ab{\sin(y - z)} &\leq \tfrac25, \\
\ab{\cos(x - y)} &\leq \tfrac25, & 
\ab{\cos(x - z)} &\leq \tfrac25, & 
\ab{\cos(y - z)} &\leq \tfrac25.
\end{align}
Let's prove this by proving the following stronger proposition: for every $x,y,z \in \mathbb R$, at least one of the following inequalities is true:
\begin{align}\newcommand{sinp}{\sin\left(\tfrac\pi8\right)}
\ab{\sin x} &\leq \sinp, & 
\ab{\sin y} &\leq \sinp, & 
\ab{\sin z} &\leq \sinp, \\
\ab{\sin(x - y)} &\leq \sinp, & 
\ab{\sin(x - z)} &\leq \sinp, & 
\ab{\sin(y - z)} &\leq \sinp,
\end{align}
The second proposition is strictly stronger than the first because
$\sinp < \tfrac25$.
The second proposition can be proved by contradiction. Suppose that for some $x,y,z \in \mathbb R$
the claim is false. Then for these particular numbers $x,y,z,$ all of the following must be true:
\begin{align}
\ab{\sin x} &> \sinp, & 
\ab{\sin y} &> \sinp, & 
\ab{\sin z} &> \sinp, \\
\ab{\cos x} &> \sinp, & 
\ab{\cos y} &> \sinp, & 
\ab{\cos z} &> \sinp, \\
\ab{\sin(x - y)} &> \sinp, & 
\ab{\sin(x - z)} &> \sinp, & 
\ab{\sin(y - z)} &> \sinp, \\
\ab{\cos(x - y)} &> \sinp, & 
\ab{\cos(x - z)} &> \sinp, & 
\ab{\cos(y - z)} &> \sinp.
\end{align}
Let $k$ be an integer and $x_0$ a real number such that $x = x_0 + \frac{k\pi}{2}$
and $0 \leq x_0 < \frac\pi2.$
Note that if $\sin x_0 \leq \sinp$ then $\ab{\sin x} \leq \sinp$.
So the assumption that the claim is false implies
$\ab{\sin x} > \sinp$,
which implies that $\sin x_0 > \sinp$,
which in turn implies that
$$ \frac\pi8 < x_0 < \frac{3\pi}{8}. $$
The same logic applies to $y$ and $z$ as well. Let $m, n$ be integers and
$y_0, z_0$ be real numbers such that
$y = y_0 + \frac{m\pi}{2}$, $0 \leq y_0 < \frac\pi2$,
$z = z_0 + \frac{n\pi}{2}$, and $0 \leq z_0 < \frac\pi2.$ Then
\begin{align}
\frac\pi8 &< y_0 < \frac{3\pi}{8}, \\
\frac\pi8 &< z_0 < \frac{3\pi}{8}.
\end{align}
Without loss of generality, we can assume $x_0 \geq y_0 \geq z_0.$
Then $x_0 - y_0$ is a real number
such that $x - y = x_0 - y_0 + \frac{(k - m)\pi}{2}$ and
$0 \leq x_0 - y_0 < \frac\pi2.$
The assumption that the claim is false then implies
$\ab{\sin(x - y)} > \sinp$,
which implies that
$\sin (x_0 - y_0) > \sinp$.
In particular,
$$ x_0 - y_0 > \frac\pi8. $$
For similar reasons,
$$ y_0 - z_0 > \frac\pi8. $$
We therefore have
$$ y_0 > z_0 + \frac\pi8 $$
and
$$
\frac{3\pi}{8} > x_0 > y_0 + \frac\pi8 > z_0 + \frac{2\pi}{8}
 > \frac{3\pi}{8},
$$
in short, $\frac{3\pi}{8} > \frac{3\pi}{8},$
which contradicts the assumption that the claim was false.
Therefore the claim is true.
A similar argument can show that for every $x,y,z \in \mathbb R$, at least one of the following inequalities is true:
\begin{align}
\ab{\cos x} &\leq \sinp, & 
\ab{\cos y} &\leq \sinp, & 
\ab{\cos z} &\leq \sinp, \\
\ab{\cos(x - y)} &\leq \sinp, & 
\ab{\cos(x - z)} &\leq \sinp, & 
\ab{\cos(y - z)} &\leq \sinp.
\end{align}
Hint: let $-\tfrac\pi2 < x_0 \leq \tfrac\pi2.$
