Inequality Question-Maximum Problem:

Let $a, b, c, d, e, f$ be nonnegative real numbers such that $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6$ and $ab + cd + ef = 3$. What is the maximum value of $a+b+c+d+e+f$?

How would I do this? Would we need to use Cauchy-Schwarz or any of those types of inequalities?
Edit: My question is different from the possible duplicate because the answers on that question are based on Lagrangian multipliers and mine is based on Cauchy-Schwarz

Use Cauchy-Schwarz on the vectors 
$$x = \begin{pmatrix}a+b \\
c+d \\
e+f\end{pmatrix}, \quad y = \begin{pmatrix}1 \\
1 \\
1\\ \end{pmatrix}$$
Then $$x\cdot x = (a+b)^2 + (c+d)^2 + (e+f)^2 \\= a^2 + b^2 + c^2 + d^2 + e^2 + f^2+ 2ab+2cd+2ef \\=6+2\times3 \\=12$$
Also, $y\cdot y = 3$, and $x \cdot y = a+b+c+d+e+f$, the quantity you're looking to maximise. Cauchy-Schwarz says $(x \cdot y)^2 \leq (x\cdot x)(y \cdot y)$, so 
$$ (a+b+c+d+e+f)^2 \leq 12 \times 3 \\a+b+c+d+e+f \leq 6$$
Also, observe that $a=b=c=d=e=f=1$ achieves this value.
