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?

Using AM-GM Inequality:
$$\frac{x+y}2\ge\sqrt{xy}\iff xy\le \left(\frac{x+y}2\right)^2$$
We have:
\begin{align}
(x+y)^3&=x^3+y^3+3xy(x+y)\\
&\le x^3+y^3+3\left(\frac{x+y}2\right)^2(x+y)\\
&=x^3+y^3+\frac 34 (x+y)^3
\end{align}
So 
$$x^3+y^3\ge \frac14 (x+y)^3$$
