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?
 A: 
Is this proof valid?

Your proof is valid and correct. However, it would be better if you prove that an equivalent statement is true, from an inequality, instead of doing the opposite.
$$\begin{split} & x^3+y^3\geq \frac{1}{4}(x+y)^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 3x^3+3y^3\geq 3xy^2+3yx^2\\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff &(x-y)^2(x+y)\geq 0 \\
\end{split}$$
Which is clearly true when $x+y \geq 0$

Is there a shorter method ?

The given inequality is equivalent to 
$$4(x^3+y^3)(x+y)=(1+1)(1+1)(x^3+y^3)(x+y) \geq (x+y)^4$$ which is true by Hölder's inequality
A: 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 $b=(1,1)$.
A: 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 
$$x^3+y^3\ge \frac14 (x+y)^3$$
A: 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.
