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Given real numbers $x,y,z$ satisfying $x+y+z=0$ and $$ \frac{x^4}{2x^2+yz}+\frac{y^4}{2y^2+zx}+\frac{z^4}{2z^2+xy}=1$$ Find all possible values of $x^4+y^4+z^4$ with proof.

My attempt : Putting $x=-y-z$ and doing subsequent calculations we get something like $$\displaystyle \sum_{cyc}\frac{x^4}{(z-x)(y-x)}=1$$ But nothing further. Can someone help me with this? Thanks in advance.

Karo
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shadow10
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  • Umm what do you mean? $x^4+y^4+z^4\ge 0$ always. And I ask for all possible values. You also need to satisfy the second equation. Please read carefully before you answer. Thanks. – shadow10 Jun 01 '14 at 06:30

2 Answers2

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"Find all possible values" ? Answer : there is only one possible value, $2$. Indeed,

$$ \begin{array}{lcl} x^4+y^4+z^4 &=& x^4+y^4+(x+y)^4 \\ &=& 2x^4+4x^3y+6x^2y^2+4xy^3+2y^4 \\ &=& 2(x^2+xy+y^2)^2 \end{array}\tag{1} $$

so it will suffice to show that $x^2+xy+y^2=1$. But

$$ 1=\frac{x^4}{2x^2+yz}+\frac{y^4}{2y^2+zx}+\frac{z^4}{2z^2+xy}= \frac{x^4}{2x^2-y(x+y)}+\frac{y^4}{2y^2-x(x+y)}+\frac{(x+y)^4}{2(x+y)^2+xy}\tag{2} $$

and a little algebra (below) shows that the rightmost expression simplifies to $x^2+xy+y^2$, so we are done.

$$\begin{array}{lcl}E &=& \frac{x^4}{2x^2+yz}+\frac{y^4}{2y^2+zx}+\frac{z^4}{2z^2+xy} \\ &=& \frac{x^4}{2x^2-y(x+y)}+\frac{y^4}{2y^2-x(x+y)}+\frac{(x+y)^4}{2(x+y)^2+xy} \\ &=& \frac{x^4}{(2x+y)(x-y)}+\frac{y^4}{(2y+x)(y-x)}+\frac{(x+y)^4}{(x+2y)(y+2x)} \\ &=& \frac{-x^4(2y+x)+y^4(2x+y)}{(2x+y)(2y+x)(y-x)}+\frac{(x+y)^4}{(x+2y)(y+2x)} \\ &=& \frac{y^5-x^5+2xy(y^3-x^3)}{(2x+y)(2y+x)(y-x)}+\frac{(x+y)^4}{(x+2y)(y+2x)} \\ &=& \frac{y^4+xy^3+x^2y^2+x^3y+x^4+2xy(y^2+xy+x^2)}{(2x+y)(2y+x)}+\frac{(x+y)^4}{(x+2y)(y+2x)} \\ &=& \frac{y^4+3xy^3+3x^2y^2+3x^3y+x^4}{(2x+y)(2y+x)}+\frac{(x+y)^4}{(x+2y)(y+2x)} \\ &=& \frac{y^4+3xy^3+3x^2y^2+3x^3y+x^4}{(2x+y)(2y+x)}+\frac{y^4+4xy^3+6x^2y^2+4x^3y+x^4}{(x+2y)(y+2x)} \\ &=& \frac{2y^4+7xy^3+9x^2y^2+7x^3y+2x^4}{(2x+y)(2y+x)} \\ &=& \frac{2y^4+7xy^3+9x^2y^2+7x^3y+2x^4}{2y^2+5xy+2x^2} \\ &=& \frac{2y^4+5xy^3+2x^2y^2}{2y^2+5xy+2x^2}+\frac{2x^4+5yx^3+2x^2y^2}{2y^2+5xy+2x^2}+ \frac{2xy^3+2x^3y+5x^2y^2}{2y^2+5xy+2x^2} \\ &=& y^2+x^2+xy \end{array}\tag{3}$$

user642796
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Ewan Delanoy
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2

this is a simple approach for $x^2+xy+y^2=1$, The op has a good start anyway, from his last step:

note:$\dfrac{x^4}{(z-x)(y-x)}=\dfrac{x^4}{y-z}\left(\dfrac{1}{z-x}-\dfrac{1}{y-x}\right)$

$\sum_{cyc}\dfrac{x^4}{(z-x)(y-x)}=\dfrac{x^4}{(z-x)(y-z)}-\dfrac{x^4}{(y-z)(y-x)}+\dfrac{y^4}{(z-y)(x-y)}+\dfrac{z^4}{(x-z)(y-z)}=\dfrac{x^4-z^4}{(z-x)(y-z)}+\dfrac{y^4-x^4}{(y-z)(y-x)}=\dfrac{-(x+z)(x^2+z^2)+(y+x)(y^2+x^2)}{y-z}=\dfrac{y(x^2+z^2)-z(y^2+x^2)}{y-z}=\dfrac{x^2(y-z)-yz(y-z)}{y-z}=x^2-yz=x^2+xy+y^2$

chenbai
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