Trigonometric inequality in a triangle If $\alpha,\beta,\gamma$ are the interior angles in a triangle, the following inequality seems to hold: 
$$1\lt\dfrac{\sin(\alpha)}{\pi-\alpha}+\dfrac{\sin(\beta)}{\pi-\beta}+\dfrac{\sin(\gamma)}{\pi-\gamma}\lt\dfrac{4}{\pi}$$
Is it possible to prove it?
 A: Note that $\gamma=\pi-\alpha-\beta$ and that $\sin(\alpha)=\sin(\pi-\alpha)$
Therefore you are asking for
$$1<\frac{\sin(\pi-\alpha)}{\pi-\alpha}+\frac{\sin(\pi-\beta)}{\pi-\beta}+ \frac{\sin(\alpha+\beta)}{\alpha+\beta}<\frac{4}{\pi}$$
Set $x=\pi-\alpha$, $y=\pi-\beta$, $z=\alpha+\beta$
you are asking for $$1<\frac{\sin x}{x}+\frac{\sin y}{y}+\frac{\sin z}{z}<\frac{4}{\pi}$$
under the constranint $x,y,z\in(0,\pi)$ and $x+y+z=2\pi$
Let $F(x,y,x)=\frac{\sin x}{x}+\frac{\sin y}{y}+\frac{\sin z}{z}$ and $f(t)=\sin(t)/t$. Note that $f$ and $f'$ are injective on $[0,\pi]$.
The gradient of $F$ is $(f'(x),f'(y),f'(z))$. By using Lagrange multipliers you see that at critical points in the interior of the domain of definition, the gradient of $F$ must be orthogonal to the plane $x+y+z=0$, that is to say $f'(x)=f'(y)=f'(z)$. Injectivity of $f'(t)$ implies $x=y=z$. 
The same argument repeated on the boundary of the domain $\{(x,y,z): x,y,z\in(0,\pi)$ and $x+y+z=2\pi\}$ tells you that
the maximum and the minimum of $F$ on such domain are atteined at one of the points
1) $x=y=z=2\pi/3$,
2) $z=\pi, x=y=\pi/2$, 
3) $z=\pi, y=\pi, x=0$
1) $F(x,y,z)=\frac{9\sqrt 3}{4\pi}\in(1,\frac{4}{\pi})$
2) $F(x,y,z)=\frac{4}{\pi}$
3) $F(x,y,z)=1$
