<s>[INST] Example 14, Chapter 1, Higher Algebra by Henry Sinclair. I stumbled across the question 14 in Higher Algebra by Henry Sinclair and S.R Knight:

If $$(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2$$
  Show $$x:a=y:b=c:z$$

I started by expanding the trinomials on both sides:
$$a^2x^2+b^2y^2+c^2z^2 + (a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = a^2x^2 + b^2y^2 + c^2z^2 + (axby + axcz) + (byax + bycz) + (czax +czby)$$
That simplifies to $$(a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = (axby + axcz) + (byax + bycz) + (czax +czby)$$
Adding like terms, I got
$$(a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = 2(axby + 2axcz + 2bycz)$$
Rearranging, I finally obtained:
$$(ay-bx)^2 + (az-cx)^2 + (bz-cy)^2 = 0$$
Now, I realized that if $\frac{x}{a} = \frac{y}{b}$ then $ay = bx$. It then follows that $ay - bx = 0$. This also applies for the other fractions.
However, in the context of this problem, we do not know that $x:a = y:b = z:c$. So, I'm not sure how I can proceed to prove it with just the equation above.
Any help would be appreciated.
 [/INST] This is the Cauchy-Schwarz inequality,$$(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$$
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