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?

In general
$$x^p + y^p \geq \dfrac{(x+y)^p}{2^{p-1}}$$
which follows from Holder's inequality, which states that
$$\Vert a \Vert_p \Vert b \Vert_q \geq \vert a\cdot b \vert$$
where $\dfrac1p + \dfrac1q = 1$. To obtain your result, take $a = (x,y)$ and $b=(1,1)$.
