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

Then prove that $a(x+y+z) = x(a+b+c)$

I did expansion on both sides and got: $a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $ but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?

Ananya
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3 Answers3

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By C-S inequality, $(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$ with equality iff $(x,y,z)=\lambda(a,b,c)$ for some $\lambda$ or $(a,b,c)=(0,0,0)$. But, if $(a,b,c)=(0,0,0)$, the problem is trivially true. If it not the case, then $x=\lambda a$, $y=\lambda b$ and $z=\lambda c$.

Then $x+y+z=\lambda(a+b+c)$. Multiplying by $a$ both sides and remembering $x=\lambda a$ yield us the proof.

sinbadh
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HINT: To do it without linear algebra, expand both sides and subtract like terms to leave

$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2acxz+2bcyz\;.$$

Notice that you can rearrange this as

$$(a^2y^2-2abxy+b^2x^2)+(a^2z^2-2acxz+c^2x^2)+(b^2z^2-2bcyz+c^2y^2)=0\;,$$

or

$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0\;.$$

  • What can you conclude about $ay-bx$, $az-cx$, and $bz-cy$?
  • What can you conclude about $a(x+y+z)$ and $x(a+b+c)$?
Brian M. Scott
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$$a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2byxz$$

$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2axcz+2byxz$$

$$a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2byxz+c^2y^2=0$$

$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$$

From here the answer is clear. All three terms are zero.

Arashium
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