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For $x,y,z\geqq 0$ and $x+y+z=1.$ Prove that$:$ $$2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \leqq \left(x^2+y^2+z^2+3xyz\right)^2.$$

Let $y=\hbox{mid} \{x,y,z\}$ and

$\text{P}= \left[\left(x^2+y^2+z^2\right)(x+y+z)+3xyz\right]^2$

$-2\left[x^2+y^2+z^2+(x+y+z)^2\right]\left[x^3y+y^3z+z^3x+xyz(x+y+z)\right]$

Ji Chen gave the following expression:

$\text{P}=\left[x^3-x^2y-xy^2-y^3+z\left(x^2-y^2+z^2-yz+zx+xy\right)\right]^2$

$+4z(x-y)(y-z)(y+z)\left(x^2+y^2+z^2+yz+zx+xy\right)\geqq 0,$

Result by SBM (me)$:$

$\text{P}=\Big[{x}^{3}-{x}^{2}y-x{y}^{2}-{y}^{3}+z \left( {x}^{2}+2\,xz-2\,yz+{z}^{2} \right) \Big]^2$

$+z\left( x-y \right) \left( y-z \right) \cdot \text{M} \geqq 0,$

where $\text{M}=\left( 2\,{x}^{3}+2\,{x}^{2}y+6\,{x}^{2}z+2\,x{y}^{2}+9\,xyz+7\,x{z}^ {2}+2\,{y}^{3}+7\,{y}^{2}z+5\,y{z}^{2}+6\,{z}^{3} \right) $

Michael Rozenberg
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NKellira
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1 Answers1

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Also, BW helps.

Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.

Thus, we need to prove that: $$\left(\sum_{cyc}(x^3+x^2y+x^2z+xyz)\right)^2\geq4\sum_{cyc}(x^2+xy)\sum_{cyc}(x^3y+x^2yz)$$ or $$16(u^2-uv+v^2)x^4+8(3u^3-5u^2v+4uv^2+3v^3)x^3+$$ $$+(17u^4-34u^3v+3u^2v^2+38uv^3+17v^4)x^2+$$ $$+2(3u^5-6u^4v-3u^3v^2+5u^2v^3+8uv^4+3v^5)x+$$ $$+(u^3-u^2v-uv^2-v^3)^2\geq0,$$ which is obvious.

Michael Rozenberg
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