<s>[INST] 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 $x\in (0,\pi]$ 
 [/INST] Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
From $\pi < \frac{22}{7}$ we deduce $\frac{3 \pi}{10} < \frac{66}{70} < 1$, and so
$$2 \sin 1 > 2 \sin \frac{3 \pi}{10} = 2 \cdot \frac{1}{4}(1 + \sqrt{5}) > \frac{8}{5}.$$ The last inequality (which itself is a reasonably tight bound on the Golden Ratio $\phi$) follows from rearranging and squaring.
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