convergence of three sequences given two conditions The problem is the following :
Given three real sequences $(u_n)$ , $v_n)$ and $(w_n)$ and a real number $a$ such that 
$$\lim\limits (u_n+v_n+w_n)=3a$$
$$\lim\limits (u_n^2+v_n^2+w_n^2)=3a^2$$
It is asked to prove that 
$$\lim u_n = \lim v_n=\lim w_n = a$$
I approached it as follows :
Let $U_:=u_n-a$, $V_n:=v_n-a$ and $W_n :=w_n-a$.
We have 
$$U_n^2 + V_n^2 + W_n^2 = (u_n^2+v_n^2+w_n^2) -2a(u_n+v_n+w_n) + 3a^2$$
So that 
$$U_n^2 + V_n^2 + W_n^2 \to 0 \quad (3)$$ 
and
$$U_n + V_n + W_n \to 0 \quad (4) $$ 
It then suffices to prove that $(U_n) $ , $(V_n)$ and $(W_n)$ converge to $0$ under assumptions $(3)$ and $(4)$ 
Given $\epsilon > 0 $ , $(3)$ implies there exists $N$ natural such that 
$$ U_n^2 \le U_n^2 + V_n^2 + W_n^2 < \epsilon$$
So that for that $N$ we have
$$|U_n| < \epsilon$$
yielding the convergence of $(U_n)$ to $0$.
We proceed similarly for $(V_n)$ And $(W_n)$.
I would like to know whether this approach is correct.
thanks
 A: Your proof is correct apart from one small point: $U_n^2 < \epsilon$ does not imply $|U_n| < \epsilon$. But that is easily fixed if you start with $$U_n^2 \le U_n^2 + V_n^2 + W_n^2 < \epsilon^2$$ instead.
Alternatively argue that 
$$
0 \le U_n^2 \le U_n^2 + V_n^2 + W_n^2 \to 0 \implies U_n^2 \to 0 \implies |U_n| \to 0
$$
 because of the “squeeze theorem” and because the square root is a continuous function on the non-negative real numbers.
That fact that $U_n + V_n + W_n \to 0 $ is actually not needed.
It is important that we have sequences of real numbers, the conclusion is not true for complex numbers. Example:
$$
 u_n = 1 \, , \, v_n = e^{2\pi i/3} \, , \, w_n = e^{4\pi i/3} 
$$
are the third roots of unity, and satisfy
$$
 u_n + v_n + w_n = 0 \, ,\\
 u_n^2 + v_n^2 + w_n^2 = 0 \, . \\
$$
A: Let $|| \cdot||$ denote the Euclidean norm on $ \mathbb R^3.$
Then we have
$||(u_n,v_n,w_n)-(a,a,a)||^2= u_n^2+v_n^2+w_n^2-2a(u_n+v_n+w_n)+3a^2 \to 3a^2-2a \cdot 3a+3a^2=0$.
Hence $(u_n,v_n,w_n)-(a,a,a) \to (0,0,0)$, thus $\lim u_n = \lim v_n=\lim w_n = a.$
