$ Ravi: x+y=a,x+z=b,y+z=c \implies a,b,c\geq 0 $ $ Then $ $ (a+b+c)^3=32abc\geq 3abc $ $ Then $ $ a,b,c\geq 0\implies Ravi $

I prooved that it is impossible. $(x+y)(y+z)(y+z)=x^2(y+z)+y^2(x+z)+z^2(x+y)+2xyz$ $x+y>z$ and $x+z>y$ and $y+z>x$ so $(x+y)(y+z)(z+x) >x^3+y^3+z^3 +(2xyz)$ That shows that the first equality is false. The answer is no.

TurtleKing123 wrote: $(x+y)(y+z)(y+z)=x^2(y+z)+y^2(x+z)+z^2(x+y)$ You're missing a $2xyz$ on the right hand side. Feridimo wrote: $ Ravi: x+y=a,x+z=b,y+z=c \implies a,b,c\geq 0 $ $ Then $ $ (a+b+c)^3=32abc\geq 3abc $ $ Then $ $ a,b,c\geq 0\implies Ravi $ That's not what we want to show. We want to show if $b+c=x,a+c=y,a+b=z$, then $a,b,c\geq 0$.

JBMO2020 wrote: Is it true that there exists a triangle with sides $x, y, z$ so that $x^3+y^3+z^3=(x+y)(y+z)(z+x)$?