If $x,y,z>0$ and $x+y+z=1$ Then prove that $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$ 
If $x,y,z$ are positive real number and $x+y+z=1\,$ Then prove that
$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$

Let $$f(x,y,z)=xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$
Then $$\frac{f(x,y,z)}{xyz} = \frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^2}{y}$$
Now Using $\bf{Cauchy-Schwarz\; }$ Inequality, We get$$\frac{f(x,y,z)}{xyz}\geq \frac{4(x+y+z)^2}{x+y+z} = 4\Rightarrow f(x,y,z)\geq 4xyz$$
My Question is How can we solve it Using $\bf{A.M\geq G.M}$ nequality,
plz explain me, Thanks
 A: First I edited your proof, you missed the factor $4$ in the answer. To use the AM-GM you can still use what you have there in the proof.
$$\dfrac{f(x,y,z)}{xyz} = \sum_{\text{cyclic}} \dfrac{(1-x)^2}{x}=\sum_{\text{cyclic}} \dfrac{1-2x+x^2}{x} = \sum_{\text{cyclic}} \dfrac{1}{x} - 6 + 1\geq \dfrac{9}{x+y+z} -6+1 = 4$$
A: Brilliant answer from @deepsea but it took me a while to understand it. For anyone else struggling here are some baby steps:
$$
\frac{f(x,y,z)}{xyz}=\frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^2}{y}
$$
$$
=\sum_{cyc}\frac{(x+y)^2}{z}=\sum_{cyc}\frac{(y+z)^2}{x}=\sum_{cyc}\frac{(1-x)^2}{x}\ \text{, because}\ 1-x=y+z
$$
$$
\sum_{cyc}\frac{(1-x)^2}{x}=\sum_{cyc}\frac{(1-2x+x^2)}{x}=\sum_{cyc}(\frac{1}{x}-2+x)
$$
$$
\sum_{cyc}\frac{1}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{(\sqrt1+\sqrt1+\sqrt1)^2}{x+y+z}\ \text{, by Titu's Lemma}
$$
$$
\frac{(\sqrt1+\sqrt1+\sqrt1)^2}{x+y+z}=\frac{3^2}{x+y+z}=\frac{9}{1}=9
$$
$$
\sum_{cyc}2=2+2+2=6\ \text{and}\ \sum_{cyc}x=x+y+z=1
$$
$$
\text{Putting it all back together,}\ \sum_{cyc}(\frac{1}{x}-2+x) \ge 9-6+1=4
$$
$$
\frac{f(x,y,z)}{xyz}\ge4\implies f(x,y,z)\ge4xyz
$$
As far as I can tell this does not depend on AM-GM. Also, I found it helpful to realise it was Titu's Lemma employed in the OP solution rather than Cauchy-Schwarz, although I appreciate one is a specialisation of the other.
