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 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.