Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$ Let $x\in R$ show that
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{8}{5}$$
since
$$f(x)=f(x+\pi),$$it sufficient to show $x\in (0,\pi]$ 

Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
Then, from Maclaurin expansion, we have
$$ \sin 1 = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \ldots $$
Observe that the absolute value of each of these terms is decreasing, hence in this alternating sum, we may conclude that
$$ \sin 1 = 1 - \frac{1}{3!} + \left( \frac{1}{5!} - \frac{1}{7!} \right) + \left( \ldots \right) + \ldots > 1 - \frac{1}{3!} = \frac{5}{6} > \frac{4}{5}. $$

In particular, we could have strengthened the original inequality to
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{5}{3}.$$
