Example 14, Chapter 1, Higher Algebra by Henry Sinclair.

Question

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.

Answer 1

If $p^2+q^2+r^2=0$ where $p,q,r\in \mathbb{R}$, then we know that $p=q=r=0$.

Hence we do know that $ay-bx=az-cx=bz-cy=0$.

Remark: This question is basically asking when does equality holds for Cauchy-Schwarz inequality.

Answer 2

This is the Cauchy-Schwarz inequality,$$(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$$