Problem. Let $x, y, z$ be real numbers with $\sin x + \sin y + \sin z = 2$ and $\cos x + \cos y + \cos z = \frac{11}{5}$. Prove that $$\sin(x + y + z) \ge \frac45.$$
This question is related to these two questions Q1, Q2.
The result can be verified by Mathematica.
In detail, denote $c_1 = \cos x, c_2 = \cos y, c_3 = \cos z, s_1 = \sin x, s_2 = \sin y, s_3 = \sin z$.
We need to prove that $s_1c_2c_3 - s_1s_2s_3 + c_1s_2c_3 + c_1c_2s_3 \ge \frac45$
given that $c_1^2 + s_1^2 = c_2^2 + s_2^2 = c_3^2 + s_3^2 = 1$ and $s_1 + s_2 + s_3 = 2$ and $c_1 + c_2 + c_3 = \frac{11}{5}$.
Mathematica says it is true.
Equality holds if e.g. $\cos x = \frac45, \cos y = \frac45, \cos z = \frac35, \sin x = \frac35, \sin y = \frac35, \sin z = \frac45$.
Is there a nice/simple proof?