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Question Number 64201 by Prithwish sen last updated on 15/Jul/19

$$\mathrm{Find}\mathrm{all}\mathrm{positive}\mathrm{solutions}$$$$\mathrm{x}+\mathrm{y}+\mathrm{z}=1$$$${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3+\mathrm{xyz}={\mathrm{x}}^4+{\mathrm{y}}^4+{\mathrm{z}}^4+1$$

Answered by MJS last updated on 15/Jul/19

$$z=1-x-y$$$$\Rightarrow$$$$y^4+2(x-1)y^3+\frac{6x^2-8x+3}{2}{y}^2+\frac{4x^3-8x^2+5x-1}{2}y+\frac{2x^4-4x^3+3x^2-x+1}{2}=0$$$$y=t-\frac{x-1}{2}$$$$t^4+\frac{x(3x-2)}{2}{t}^2+\frac{9x^4-12x^3+4x^2+7}{16}=0$$$$\Rightarrow t^2=-\frac{x(3x-2)}{4}\pm \frac{\sqrt{7}}{4}\mathrm{i}$$$$\Rightarrow \mathrm{no}\mathrm{real}\mathrm{solutions}\mathrm{at}\mathrm{all}$$

Commented by MJS last updated on 15/Jul/19

$$\mathrm{the}\mathrm{maximum}\mathrm{of}$$$$-x^4-y^4-z^4-1+x^3+y^3+z^3+xyz$$$$\mathrm{is}-\frac{7}{8}$$

Commented by Prithwish sen last updated on 15/Jul/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by ajfour last updated on 16/Jul/19

$${(x+y+z)}^4=x^4+{(y+z)}^4+4x^3(y+z)$$$$+6x^2{(y+z)}^2+4x{(y+z)}^3$$$$\Rightarrow x^4+y^4+z^4+4yz[{(y+z)}^2-2yz)$$$$+6y^2{z}^2+6x^2{(y+z)}^2+4x{(y+z)}^3=1$$$$$$$$orsincey+z=1-x$$$$\Rightarrow$$$$x^4+y^4+z^4-2y^2{z}^2+6x^2{(1-x)}^2$$$$+4x{(1-x)}^3=1$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow 2y^2{z}^2=x^4+y^4+z^4+(2x^2+4x){(1-x)}^2-1$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$......\left(i\right)$$$${(x+y+z)}^3=x^3+y^3+z^3$$$$+3[x^2(y+z)+y^2(z+x)+z^2(x+y)]$$$$+6xyz$$$$\Rightarrow x^3+y^3+z^3+3x^2(1-x)+3yz(1-x)$$$$+3x(1-2yz)+6xyz=1$$$$\Rightarrow yz=\frac{x^3+y^3+z^3-3x^3+3x^2+3x-1}{3x-3}$$$$or$$$$3(x-1)yz=x^3+y^3+z^3+(1-x^2)(3x-1)$$$$.......\left(ii\right)$$$$\left(i\right)-\left(ii\right)$$$$2y^2{z}^2-3(x-1)yz=xyz-1$$$$+2x(x+2){(1-x)}^2-1-(1-x^2)(3x-1)$$$$\Rightarrow$$$$2yz(yz+x)+2=(1-x)[-2x^3-2x^2+4x-3x^2-2x+1)$$$$\Rightarrow 2yz(yz+x)+2+$$$$(1-x)(2x^3+5x^2-2x-1)=0$$$$say$$$$y^2{z}^2+x\left(yz\right)+1+f=0$$$$yz=-\frac{x}{2}+\sqrt{\frac{x^2}{4}-(1+f)}$$$$andy+z=1-x$$$$\Rightarrow$$$$y,zarerootsofeq.$$$$s^2-(1-x)s+\sqrt{\frac{x^2}{4}-(1+f)}-\frac{x}{2}=0$$$$\Rightarrow y,z=\frac{1-x}{2}\pm \sqrt{\frac{{(1-x)}^2}{4}+\frac{x}{2}-\sqrt{\frac{x^2}{4}-(1+f)}}$$$$for+veyandz,x<1$$$$1+f<0$$$$2+2f=(1-x)(2x^3+5x^2-2x-1)+2<0$$$$\Rightarrow 2x^4+3x^3-7x^2+x-1>0$$$$forx\in (0,1)$$$$butfornorealxisthistrue,$$$$\left(seegraphbelow\right)..$$

Commented by ajfour last updated on 16/Jul/19

Commented by Prithwish sen last updated on 16/Jul/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

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