Question: Consider:

$$\lim\limits\_{n\to +\infty}\sqrt[n]{|\sin(n+1)|+|\sin(n)|}=1$$

I figured out that
$$\lim\limits\_{n\to +\infty}\sqrt[n]{|\sin(n+1)|+|\sin(n)|}\leqslant\lim\_{n\to +\infty}\sqrt[n]{2}=1$$

What about the other side inequality?

Answer: 
You have $2>|\sin(n+1)|+|\sin(n)|>0$ is continuous and periodic, so there is an $a$ witch satisfies $a>0$ and $2>|\sin(n+1)|+|\sin(n)|>a$, now you can take the n-th root of a, bacause $\lim\_{n\to +\infty}\sqrt[n]{a}=1$
It is enough to study $f(x)=\left|\sin(x)\right|+\left|\sin(x+1)\right|$ over $[0,\pi]$ to be able to state that:
$$\forall x\in\mathbb{R},\qquad \left|\sin(x)\right|+\left|\sin(x+1)\right|\geq \sin(1). $$
Equality is attained at $x\in\pi\mathbb{Z}$ and at $x\in -1+\pi\mathbb{Z}$. 

We may also improve the upper bound in the following way:
$$\forall x\in\mathbb{R},\qquad \left|\sin(x)\right|+\left|\sin(x+1)\right|\leq 2\cos\left(\frac{1}{2}\right). $$
Equality is attained at $x=\frac{\pi-1}{2}+\pi\mathbb{Z}$.