<s>[INST] 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
 [/INST] 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.
$$\dfrac{f(x,y,z)}{xyz} = \sum_{\text{cyclic}} \dfrac{(1-x)^2}{x}=\sum_{\text{cyclic}} \dfrac{1-2x+x^2}{x} = \sum_{\text{cyclic}} \dfrac{1}{x} - 6 + 1\geq \dfrac{9}{x+y+z} -6+1 = 4$$
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