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Let $a , b \in \mathbb R$. Then $ | a^3 + a^2 b + ab^2 + b^3 | \leqslant |a + b|^3 $ holds?

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    If you require $a,b\geq 0$ then yes. Otherwise, no.2012-07-01

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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 $a,b$.

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    Be my guest:), maybe you will reach something intersting:)2012-07-01
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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 $RHS-LHS\ge 0$: $|a+b|\cdot 2ab\ge 0$ $a=-b\quad\text{or}\quad ab\ge 0$

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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.