Find the value of $\frac{a+b+c}{d+e+f}$ Given real numbers $a, b, c, d, e, f$, such that:
$a^2 + b^2 + c^2 = 25$
$d^2 + e^2 + f^2 = 36$
$ad + be + cf = 30$
What is the value of $\frac{a+b+c}{d+e+f}$?
I've tried combining equations in several ways but haven't gotten very far. Any hints would be appreciated.
 A: Hint:   by the Cauchy-Schwarz inequality:
$$
900 = 30^2 = (ad + be + cf )^2 \le (a^2 + b^2 + c^2 )(d^2 + e^2 + f^2) = 25 \cdot 36 = 900
$$
Equality occurs when $a,b,c$ and $d,e,f$ are proportional.
A: Let $\vec u = (a,b,c)$ and $\vec v = (d,e,f)$.
Your equations tell you that $\lVert \vec u \rVert = 5$, $\lVert \vec v \rVert = 6$ and $\vec u \cdot \vec v = 30$. The formula for dot products tells you that
$$30 = \vec u \cdot \vec v = \lVert \vec u \rVert \lVert \vec v \rVert \cos \theta = 30\cos\theta$$
where $\theta$ is the (acute) angle between $\vec u$ and $\vec v$, so that $\cos \theta = 1$ and hence $\theta = 0$.
Hence $\vec u$ and $\vec v$ are parallel.
Since they're parallel, they have the same angle, $\varphi$ say, with the vector $(1,1,1)$. Finally, we have
$$\frac{a+b+c}{d+e+f} = \frac{\vec u \cdot (1,1,1)}{\vec v \cdot (1,1,1)} = \frac{\lVert \vec u \rVert \lVert (1,1,1) \rVert \cos \varphi}{\lVert \vec v \rVert \lVert (1,1,1) \rVert \cos \varphi} = \frac{\lVert \vec u \rVert}{\lVert \vec v \rVert} = \frac{5}{6}$$
A: By C-S $$(a^2+b^2+c^2)(d^2+e^2+f^2)\geq(ad+be+cf)^2,$$
where the equality occurs for $(a,b,c)||(d,e,f)$, which we got.
Let $(a,b,c)=k(d,e,f).$ 
Thus, $5=|k|6$ and $|k|=\frac{5}{6}$ and
$$\frac{a+b+c}{d+e+f}=\frac{5}{6}$$ or
 $$\frac{a+b+c}{d+e+f}=-\frac{5}{6}$$
