Is this inequality true $xy(x^{2}+y^{2})\le\frac{(x+y)^{4}}{8}$?

Question

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

Answer 1

It it equivalent to $$\frac{1}{8} (x-y)^4\geq 0$$ and this is true for all $$x,y\in\mathbb{R}$$

Answer 2

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 $xy(x^2 + y^2) \le \frac {(x+y)^4}8\iff$

$(x+y)^4 - 8x^3y - 8xy^3 \ge 0 \iff$

$(x^4 + 4x^3y +6x^2y^2 + 4xy^3 + y^4) - 8x^3y - 8xy^3 \ge 0 \iff$

$x^4 - 4x^3y +6x^2y^2 - 4xy^3 + y^4 \ge 0 \iff$

$(x-y)^4 \ge 0$.

Answer 3

By AM-GM: $$\frac{(x+y)^4}{8}=\frac{(x^2+y^2+2xy)^2}{8}\geq\frac{\left(2\sqrt{(x^2+y^2)\cdot2xy}\right)^2}{8}=xy(x^2+y^2).$$