# Prove $\left(\dfrac{x^3+y^3}{2}\right)^2\geqslant\left(\dfrac{x^2+y^2}{2}\right)^3$

I wanted some help proving this inequality ($$x$$ and $$y$$ are positive constants):

$$\left(\dfrac{x^3+y^3}{2}\right)^2\geq\left(\dfrac{x^2+y^2}{2}\right)^3$$

and possibly extend this to $$\left(\dfrac{x^3+y^3+z^3}{3}\right)^2\geq\left(\dfrac{x^2+y^2+z^2}{3}\right)^3$$

where $$z$$ is also positive. It's possible I'm just having a mind-blank. What I've tried so far:

• Showing LHS-RHS is greater than or equal to 0, but I'm unsure of the steps afterwards to deal with the negative terms
• Letting $$f(x) = \left(\dfrac{x^3+a^3}{2}\right)^2-\left(\dfrac{x^2+a^2}{2}\right)^3$$ for constant positive $$a$$ and use calculus. However it's a bit extreme for the topic I'm studying right now which focuses on elementary algebraic techniques to prove these inequalities (e.g basic things like AM-GM).

I also suppose this inequality looks quite common, but I don't know, how to look it up - if it's possible to direct me towards a similar thread that would be great.

• Yes, was a typo on my behalf - apologies! Jul 19, 2017 at 11:56
• This is the power mean inequality : en.wikipedia.org/wiki/Generalized_mean. Jul 19, 2017 at 11:56
• why don't you just expand it out? Jul 19, 2017 at 12:24

you have $$\left(\frac{x^3+y^3}{2}\right)^3-\left(\frac{x^2+y^2}{2}\right)^3=1/8\, \left( {x}^{4}+2\,y{x}^{3}+2\,x{y}^{3}+{y}^{4} \right) \left( - y+x \right) ^{2} \geq 0$$
Let $x^2=a$, $y^2=b$ and $z^2=c$.
Hence, we need to prove that $$\frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}}{3}\geq\left(\frac{a+b+c}{3}\right)^{\frac{3}{2}},$$ which is Jensen for $f(x)=x^{\frac{3}{2}}$.