Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$ From R. D. Sharma's Objective Mathematics,

Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$


My attempt:
By A.M.-G.M. inequality,
$$\begin{align}
  (x+y)^2 &\geq  4(xy) \\
  (y+z)^2 &\geq  4(yz) \\
  (z+x)^2 &\geq  4(zx)
\end{align}$$
By multiplying by $xy$, $yz$, $zx$ in these equations respectively, gives
$$\begin{align}
 xy(x+y)^2 &\geq 4(xy)^2 \\
 yz(y+z)^2 &\geq 4(yz)^2 \\
 zx(z+x)^2 &\geq 4(zx)^2 \\
\end{align}$$
Adding these equations, we get
$$  A \geq 4[(xy)^2 + (yz)^2 + (zx)^2] $$
Using AM GM inequality once again on RHS, we get
$$ A \geq 12(xyz)^{\tfrac43}$$
However, provided answer key says that minimum value is $4xyz$. I do not want a solution but only wants to know that is  something wrong in my procedure? Why?
EDIT : This question has been identified as a possible duplicate of another question . However, readers will believe that my problem is different if they re-read the paragraph just above .
 A: Note that,
If $x,y\to 0$ with $z\to 1$, we have
$$\inf \left\{xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\mid x+y+z=1 \wedge x>0\wedge y>0\wedge z>0\right\}=0$$
As for obtaining the inequality $A≥4xyz$, you can easily obtain this result using the Cauchy-Schwarz inequality:
$$(xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2)\left(\frac 1{xy}+\frac 1{yz}+\frac 1{xz}\right)≥4(x+y+z)^2=4$$
$$\frac A{xyz}≥4\implies A≥4xyz.$$
But, note that $xyz$ is not a constant. Therefore, $4xyz$ is not a "minimum". We only proved that,
$$xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2≥4xyz$$
The inequality you want to prove.
A: You have not used the condition of $x+y+z=1$. Since when $x=y=z=1/3$ the equality holds, your answer is the same with the $4xyz$ answer. So the textbook answer is flawed. you can get a numerical minimum value of $4/27$.
A: An approach with symmetric functions:
$$A=xy(1-z)^2+xz(1-y)^2+yz(1-x)^2\\
\quad =xy+xz+yz-6xyz+xyz(x+y+z)\\
\quad =xy+xz+yz-5xyz$$
the symmetric $\quad xyz\quad$ in terms of $\quad x^3+y^3+z^3:$
$(x+y+z)^3=p_3+3(1-x)(1-y)(1-z)=\cdots \\
xyz=\frac 1 3(x^3+y^3+z^3)+(xy+xz+yz)-\frac 1 3$
note also $\quad xy+xz+yz=\frac 1 2\left[(x+y+z)^2-(x^2+y^2+z^2)\right]$
Then
$$A=xy+xz+yz-5xyz\\
=2(x^2+y^2+z^2)-\frac 5 3 (x^3+y^3+z^3)-\frac 1 3\\
\ge 2 \left(\frac 1 3\right)-\frac 5 3\left(\frac 1 9 \right)-\frac 1 3=\frac 4{27}$$
Given that $$\begin{aligned}\min (x^2+y^2+z^2)=\frac 1 3\\
\min(x^3+y^3+z^3)=\frac 1 9\end{aligned}$$
