# 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

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. $$\dfrac{f(x,y,z)}{xyz} = \sum_{\text{cyclic}} \dfrac{(1-x)^2}{x}=\sum_{\text{cyclic}} \dfrac{1-2x+x^2}{x} = \sum_{\text{cyclic}} \dfrac{1}{x} - 6 + 1\geq \dfrac{9}{x+y+z} -6+1 = 4$$