Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ 
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.

Let's look at 
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff & x^3+y^3\geq \frac{1}{4}(x+y)^3
\end{split}$$
is the proof valid? is there a shorter way?

Write $x=m+z$ and $y=m-z$, so that $x+y=2m\gt0$.  Then
$$x^3+y^3=(m+z)^3+(m-z)^3=2m^3+6mz^2\ge2m^3={1\over4}(2m)^3={1\over4}(x+y)^3$$
Note that we only need for the midpoint $m$ between $x$ and $y$ to be positive in order for the inequality to hold.
