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.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ $\{a,b,c,d\} \in \mathbb{R}$? $\endgroup$ – Maazul May 20 '13 at 13:20
-
2$\begingroup$ @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
Add a comment
|
$\begingroup$
$\endgroup$
1
Hint $a^3+b^3+c^3\ge 3abc$ for $a, b, c \ge 0$.
$\begingroup$
$\endgroup$
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.
