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.
 A: 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.
A: 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 $x,y>0$.
A: 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
$$\left(\frac{x^3+y^3}{2}\right)^\frac13≥\left(\frac{x^2+y^2}{2}\right)^\frac12\iff \left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$$
A: Another longer way...
Let $x=r\cos\theta$ and $y=r\sin\theta$ where $0<\theta<\frac\pi 2$ and $r>0$. Then the expression becomes
$$|\sin(2\theta)-2||\sin(\theta+\frac\pi4)|\geq\frac{1}{\sqrt{2}}$$
which is true for $0<\theta<\frac\pi 2$.
