An inequality for a sum of sine functions

Question

I want to show for $x\in\mathbb{R}$, the following inequality holds. $$ \left|\sin x\right|+\left|\sin (x+1)\right|+\left|\sin (x+2)\right|>8/5. $$ This trigonometric inequality has been verified by Mathematica using the Plot commend. However, using the sum formula for the sine function seems doesn't work, and I cannot give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks!

Answer 1

Since $f:x\mapsto \left|\sin x\right|+\left|\sin (x+1)\right|+\left|\sin (x+2)\right|$ is $\pi$-periodic, it suffices to prove the inequality for $x\in [0,\pi]$.

Also note that $$\begin{cases} f(x) = \sin(x)+\sin(x+1)+\sin(x+2) & \text{if } x\in [0,\pi-2] \\ f(x)= \sin(x)+\sin(x+1)-\sin(x+2) & \text{if } x\in [\pi-2,\pi-1] \\ f(x) = \sin(x)-\sin(x+1)-\sin(x+2) & \text{if } x\in [\pi-1,\pi] \end{cases}$$

Note that each of the subfunction is concave on each of the relevant interval. For example, if you let $f_2(x)=\sin(x)+\sin(x+1)-\sin(x+2)$, $$f_2''(x)=-f_2(x)\leq 0 \quad \text{when } x\in [\pi-2,\pi-1]$$ The argument is similar for the other functions. Therefore, each subfunction reaches its minimum at the boundary of its domain, hence $f$ reaches a minimum which is $$\min(f(0),f(\pi-2),f(\pi-1)) $$

Note that $$f(0)=f(\pi-2)= \sin(1)+\sin(2)\sim1.75$$ and $$f(\pi-1)=2\sin (1)\sim 1.68$$

The minimum is therefore $2\sin( 1)\sim 1.68 > \frac 85$