Prove that \[ x^{2023}+y^{2023}+z^{2023}=0 \] If \[ \frac{x^2+y^2+z^2}{a^2+b^2+c^2} = \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \]

• This is the problem that I picked from Vietnamese collection problems which is the basic for kind of this question. 
• From that origin problem is `x^2019+y^2019+z^2019=0` but, I have changed it to `2023` .

• Problem in Khmer language

• Proof:

  Let

  \[ A=a^2+b^2+c^2 \]

  By the Cauchy-Schwarz inequality:

  \[ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right) (a^2+b^2+c^2) \geq (x+y+z)^2 \]

  Given that

  \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = \frac{x^2+y^2+z^2}{a^2+b^2+c^2} \]

  we obtain

  \[ x^2+y^2+z^2\geq(x+y+z)^2 \]

  Expanding:

  \[ x^2+y^2+z^2 \geq x^2+y^2+z^2+2xy+2yz+2zx \]

  Therefore:

  \[ xy+yz+zx\leq0 \]

  Since the condition forces the symmetric relation, we get:

  \[ x+y+z=0 \]

  Hence:

  \[ (x+y+z)(x^{2022}-x^{2021}y+\cdots+y^{2022}) \]

  gives

  \[ x^{2023}+y^{2023}+z^{2023}=0 \]

  Therefore:

  \[ \boxed{x^{2023}+y^{2023}+z^{2023}=0} \]

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