<s>[INST] Show that if $x,y>0$, $\left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$ Through some rearrangement of the inequality and expansion, I have been able to show that the inequality is equivalent to
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6≥0$$
However, I am not sure how to prove the above or if expansion and rearrangement are even correct steps.
 [/INST] Note that your inequality is a trivial equality for $x=y$ i.e. $x^6=x^6$ as the factors $2$ cancel out.
This indicates that you can factorize $(x-y)$ out of your final equation (i.e. it's value is $0$ when $x=y$)
You get:
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6
=(x-y)(x^5+x^4y-2x^3y^2+2x^2y^3-xy^4-y^5)$$
The degree $5$ RHS expression again is zero when $x=y$, this can be seen because very symmetric in $x,y$ and signs alternate for quantities such as $+x^4y$ and $-xy^4$, so they cancel out.
So we get to factorize one more time:
$$\underbrace{(x-y)^2}_{\ge 0}\ \underbrace{(x^4+2x^3y+2xy^3+y^4)}_{>0}$$
Now the degree $4$ RHS has only $+$ signs so the expression is strictly positive considering $x,y>0$.
</s>