Is this inequality true $xy(x^{2}+y^{2})\le\frac{(x+y)^{4}}{8}$? Is this inequality true ? 
$$xy(x^{2}+y^{2})≤\dfrac{(x+y)^{4}}{8}$$ 
$x,y>0$ 
If true how ? And which inequality has use it ?
I know that : $xy≤\dfrac{(x+y)^{2}}{4}$ by Am-Gm
But is $x^{2}+y^{2}≤\dfrac{(x+y)^{2}}{2}$ ? 
In first in this rule : $\dfrac{x^{n}+y^{n}}{2}≥\left(\dfrac{x+y}{2}\right)^{n}$ 
But we have $≤$ not $≥$ 
Also after simplified a get : 
$$2x^{2}+2y^{2}=x^{2}+y^{2}+2xy$$
$$x^{2}+y^{2}-2xy=(x-y)^{2}≥0$$
Correct $\color{#2f0}{\checkmark}$ 
Now I need generalized $$x^{p}+y^{p}≤\dfrac{(x+y)^{p}}{p}$$ 
I'm correct or no ? And where deferent between this inequality and power mean inequality 

Noodling.
I see $xy$ and I see $x^2 +y^2$ and think somehow I want to convert $xy(x^2 + y^2) \to x^2 + 2xy + y^2$ but gob only knows how.
I've got a hammer called AM.GM so I whack $(x+y)^2 = x^2 + 2xy + y^2$ with it just to see what will happen.  
$2xy + x^2 + y^2 \ge  2\sqrt{2xy(x^2 + y^2)}$ happens
and well.... I can work with that.
$(x+y)^2 \ge 2\sqrt{2xy(x^2 + y^2)}$ so
$(x+y)^4 \ge 2^2\cdot [2xy(x^2 + y^2)]=8xy(x^2 + y^2)$ and .... that does it.
.....
Alternatively $xy(x^2 + y^2) \le \frac {(x+y)^4}8\iff$
$(x+y)^4 - 8x^3y - 8xy^3 \ge 0 \iff$
$(x^4 + 4x^3y +6x^2y^2 + 4xy^3 + y^4)  - 8x^3y - 8xy^3 \ge 0 \iff$
$x^4 - 4x^3y +6x^2y^2 - 4xy^3 + y^4 \ge 0 \iff$
$(x-y)^4 \ge 0$.
