Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$

Since $\sin$ is a concave function on $[0,\pi]$ and sum of concave functions is a concave function,

we have $$\min_{[0,\pi]}f=\min\{f(0),f(\pi-1),f(\pi-2),f(\pi)\}=f(\pi-1)=2\sin1>\frac{8}{5}$$

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.

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

$$ \sin 1 = 1 - \frac{1}{3!} + \left( \frac{1}{5!} - \frac{1}{7!} \right) + \left( \ldots \right) + \ldots > 1 - \frac{1}{3!} = \frac{5}{6} > \frac{4}{5}. $$

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In particular, we could have strengthened the original inequality to

$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{5}{3}.$$