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I was solving an inequality and got stuck at this part: $$abc+abd+acd+bcd\le a^3+b^3+c^3+d^3$$ Why is this true? I think it has a similar solution as $ab+ba\le a^2+b^2$, because in both cases the left side is rearranged.

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  • $\begingroup$ $\{a,b,c,d\} \in \mathbb{R}$? $\endgroup$ – Maazul May 20 '13 at 13:20
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    $\begingroup$ @Maazul: It has to be $a,b,c,d\geq 0$, because otherwise you could replace each value with its negative, and this would reverse the inequality. $\endgroup$ – Glen O May 20 '13 at 13:24
  • $\begingroup$ oh yes, sorry, I forgot to say this. $\endgroup$ – D180 May 20 '13 at 13:36
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Hint $a^3+b^3+c^3\ge 3abc$ for $a, b, c \ge 0$.

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  • $\begingroup$ Yes, I got it. Thank you :) $\endgroup$ – D180 May 20 '13 at 13:39
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Just another track to the truth: Apply $3^\text{rd}$ degree AM-GM $$abc\:\leqslant\:\frac{a^3+b^3+c^3}{3}$$ to each summand on the LHS, followed by 'garbage collection'.

Still another track {w,c}ould exploit the Rearrangement inequality.

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