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.

Use that for positive $a$ and $b$
$$a^2\ge b^3\iff \left(a^2\right)^\frac16\ge \left(b^3\right)^\frac16\iff a^{\frac13}\ge b^{\frac12}$$
then refer to the generalized mean inequality that is
$$\left(\frac{x^3+y^3}{2}\right)^\frac13≥\left(\frac{x^2+y^2}{2}\right)^\frac12\iff \left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$$
