By AM-GM $$\frac{(x+y)^3}{z}+4z^2=\frac{(x+y)^3}{2z}+\frac{(x+y)^3}{2z}+4z^2\ge3\sqrt{\frac{(x+y)^6}{4z^2}\cdot 4z^2}=3(x+y)^2$$Using similar inequalities $$\frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y}\ge 3(x+y)^2+3(y+z)^2+3(z+x)^2-4x^2-4y^2-4z^2=2(x^2+y^2+z^2)+6(xy+yz+zx)$$So we want to prove $$2(x^2+y^2+z^2)+9xyz\ge 3(xy+yz+zx)$$Because $x + y + z = 1$ holds this is equivalent to $$2(x^2+y^2+z^2)(x + y + z)+9xyz\ge 3(xy+yz+zx)(x + y + z)\iff$$$$2(x^3+y^3+z^3+xy(x+y)+yz(y+z)+zx(z+x))+9xyz\ge 3(xy(x+y)+yz(y+z)+zx(z+x)+3xyz)\iff$$$$2(x^3+y^3+z^3)\ge xy(x+y)+yz(y+z)+zx(z+x)$$This is true by many inequalities: Muirhead, twice used rearrangement or AM-GM as below: (!)$$x^3+y^3=\frac{x^3+2y^3}{3}+\frac{y^3+2x^3}{3}\ge xy^2+y^2x=xy(x+y)$$and cyclic inequalities Our step (!) gives that necessary condition for equality is $x=y=z=\frac{1}{3}$. If you plug this into the initial inequality you'll se that this is also sufficient condition, so equality iff $x=y=z=\frac{1}{3}$.

Dedicated to all fans of tangent lines. Because $z\in(0;1)$ $$0\ge (3z-1)^2(z-9)\iff\frac{(x+y)^3}{z}=\frac{(1-z)^3}{z}\ge \frac{28}{9}-\frac{20}{3}\cdot z$$Using similar inequalities and $x+y+z=1$ we are left to prove $$8+27xyz\ge 27(xy+yz+zx)$$which under given equation is equivalent to $$8(x+y+z)^3+27xyz\ge 27(xy+yz+zx)(x+y+z)\iff$$$$8(x^3+y^3+z^3)\ge 3xy(x+y)+3yz(y+z)+3zx(z+x)+6xyz$$which is true by AM-GM used like in previous post: $$2(x^3+y^3+z^3)\ge xy(x+y)+yz(y+z)+zx(z+x)$$and $$x^3+y^3+z^3\ge 3xyz$$

Lukaluce wrote: Let $x$, $y$, and $z$ be positive real numbers such that $x + y + z = 1$. Prove that $$\frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} + 9xyz \ge 9(xy + yz + zx).$$ When does equality hold? Let $x$, $y$, and $z$ be positive real numbers such that $x + y + z = 1$. Prove that $$\frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} + 9xyz \ge 9(x^2+y^2+z^2)$$

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This is a big bump, but here's the official solution: $$\frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} + 9xyz \ge 3\sqrt[3]{\frac{(x+y)^3}{z}\cdot\frac{(y+z)^3}{x}\cdot\frac{(z+x)^3}{y}}+9xyz= 3\frac{(x+y)(y+z)(z+x)}{\sqrt[3]{xyz}}+9xyz\ge 3\frac{(x+y)(y+z)(z+x)}{\frac{x+y+z}{3}}+9xyz=9((x+y)(y+z)(z+x)+xyz)=9(x+y+z)(xy+yz+zx)=9(xy + yz + zx)$$

Notice: LHS $\ge\frac{3(x+y)(y+z)(z+x)}{\sqrt[3]{xyz}}+9xyz$ $\Rightarrow$ $\frac{3\prod{x+y}}{\sqrt[3]{xyz}}$ $\ge9\prod{x+y}$ Thus we need to prove that $\frac{3}{\sqrt[3]{xyz}}$ $\ge9$ $\Longrightarrow$ $1\ge3\sqrt[3]{xyz}$ But we know that $x+y+z=1$ thus $x+y+z\ge3\sqrt[3]{xyz}$ which is trivial by AM-GM. And we are done!