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.

Since $f(x)=x^\frac{3}{2}$ is convex in $\mathbb{R}^+$ (its second derivative is positive), we have
$$\left(\frac{x^2+y^2}{2}\right)^{3/2}=f\left(\frac{x^2+y^2}{2}\right)\leq\frac{f(x^2)+f(y^2)}{2}=\frac{x^3+y^3}{2}.$$
by the definition of convexity. Taking the square of both sides gives the result.
