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: 
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 $x=\frac{\pi-1}{2}+\pi\mathbb{Z}$.
It is known that the zeroes of $\sin$ are at the point $n\pi$ for integer $n$. In particular, $\sin x$ and $\sin(x+1)$ cannot both be zero, that is, $|\sin x| + |\sin (x+1)| > 0$. The function $f(x) = |\sin x| + |\sin (x+1)|$ is positive and periodic with period $2\pi$, so due to compactness it attains its minimum somewhere in $[0,2\pi]$, let's say at the point $x\_0$. Thus $f(x) \geq f(x\_0) > 0$ for all $x$, and the limit follows.