# 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.

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 $$x,y>0$$.

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

$$\left(\frac{x^3+y^3}{2}\right)^\frac13≥\left(\frac{x^2+y^2}{2}\right)^\frac12\iff \left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$$

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 $$0<\theta<\frac\pi 2$$.