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 $x\in (0,\pi]$ 
 A: 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.
A: 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
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{5}{3}.$$
A: We'll prove that
$$\sin1>\frac{4}{5}$$ by hand.
Indeed, let $f(x)=\sin{x}-x+\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040},$ where $x\in\left[0,\frac{\pi}{2}\right].$
Thus, $$f'(x)=\cos{x}-1+\frac{x^2}{2}-\frac{x^4}{24}+\frac{x^6}{720};$$
$$f''(x)=-\sin{x}+x-\frac{x^3}{6}+\frac{x^5}{120};$$
$$f'''(x)=-\cos{x}+1-\frac{x^2}{2}+\frac{x^4}{24};$$
$$f''''(x)=\sin{x}-x+\frac{x^3}{6}$$ and 
$$f'''''(x)=\cos{x}-1+\frac{x^2}{2}=\frac{x^2}{2}-2\sin^2\frac{x}{2}=2\left(\frac{x}{2}-\sin\frac{x}{2}\right)\left(\frac{x}{2}+\sin\frac{x}{2}\right)\geq0,$$
which since
$$f''''(0)=f'''(0)=f''(0)=f'(0)=0,$$ says that $f$ increases.
Thus, $$f(1)>0$$ or $$\sin1-1+\frac{1^3}{6}-\frac{1^5}{120}+\frac{1^7}{5040}>0$$ or
$$\sin1>\frac{4241}{5040}$$ and since
$$\frac{4241}{5040}>\frac{4}{5},$$ we are done! 
A: Since $\sin$ is a concave function on $[0,\pi]$ and sum of concave functions is a concave function, 
we have
$$\min_{[0,\pi]}f=\min\{f(0),f(\pi-1),f(\pi-2),f(\pi)\}=f(\pi-1)=2\sin1>\frac{8}{5}$$
