$ | a^3 + ab^2 + a^2 b + b^3 | \leqslant |a + b|^3 $ holds?? Let $a , b \in \mathbb R$. Then $$ | a^3 + a^2 b + ab^2 + b^3 | \leqslant |a + b|^3 $$ holds?
 A: No it does not take $ a=2$ and $b=-1$ and you get $ 5 \leq 1$
EDIT:
Also for better studying this inequality suppose $ b \not = 0$ and divide with $ |b|^3$ so you get the following inequality $|x^3 + x^2 +x+1| \leq |x+1|^3$ and then for each value it holds you take a corresponding line of the plane. Also note you could do that because the inequality was homogenous for the degrees of $a,b$.
A: The left hand side is:
$$|a^3+a^2b+ab^2+b^3|=|(a+b)(a^2+b^2)|=|a+b|\cdot(a^2+b^2)$$
The right hand side is:
$$|a+b|^3=|a+b|\cdot(a+b)^2=|a+b|\cdot(a^2+2ab+b^2)$$
So the inequality is true if and only if $RHS-LHS\ge 0$:
$$|a+b|\cdot 2ab\ge 0$$
$$a=-b\quad\text{or}\quad ab\ge 0$$
A: Perhaps that would be a proof in the positive case : 
$$
| a^3 + a^2b + ab^2 + b^3 | = a^3 + a^2b + ab^2 + b^3 \le a^3 + 3a^2b + 3ab^2 + b^3  = (a+b)^3 = |a+b|^3
$$
But this works only in the case $a,b \ge 0$. In general though you don't expect this to be true, so that's (maybe) as general as one could guess.
EDIT: Note that it is also true in the case $a = -b$, so it would be possible to strive for a larger domain where this identity holds in general. Mathematica or Sage would also help conjecturing if one would plot $f(x,y) = |x+y|^3 - |x^3 + x^2y + xy^2 + y^3|$ and see where it is positive. If you are really interested in knowing when this inequality holds you should do this as your first step.
