TITLE: Simpler proof for $\frac{a^3b}{c}+\frac{b^3c}{d}+\frac{c^3d}{a}+\frac{d^3a}{b}\geq a^3+b^3+c^3+d^3$
QUESTION [5 upvotes]: Let $a\geq b\geq c\geq d>0$. Prove that:
$$\frac{a^3b}{c}+\frac{b^3c}{d}+\frac{c^3d}{a}+\frac{d^3a}{b}\geq a^3+b^3+c^3+d^3$$
I have a proof, but my proof is very ugly:
Let $c=d+u$, $b=d+u+v$ and $a=d+u+v+w$, where $u$, $v$ and $w$ are non-negatives.
After these substitutions we'll get something obvious.
Maybe there is something nice?
Thank you!

REPLY [4 votes]: Make the change: $x_1 \ge x_2 \ge x_3 \ge x_4$
$$a=e^{x_1},b=e^{x_2},c=e^{x_3},d=e^{x_4}$$
Then the inequality re written as:
$$e^{3x_1+x_2-x_3}+e^{3x_2+x_3-x_4}+e^{3x_3+x_4-x_1}+e^{3x_4+x_1-x_2} \ge e^{3x_1}+e^{3x_2}+e^{3x_3}+e^{3x_4}$$
$$A=(3x_1+x_2-x_3, 3x_2+x_3-x_4,3x_4+x_1-x_2,3x_3+x_4-x_1)$$
$$B=(3x_1, 3x_2,3x_4,3x_3)$$
Set $A$ majorizes set $B$ $(A \succ B):$
$$A_1 \ge B_1$$
$$A_1+A_2 \ge B_1+B_2$$
$$A_1+A_2+A_3 \ge B_1+B_2+B_3$$
$$A_1+A_2+A_3+A_4=B_1+B_2+B_3+B_4$$
$f(x)=e^x -$ concave function, so by Karamata's_inequality:
$$f(A_1)+f(A_2)+f(A_3)+f(A_4)\ge f(B_1)+f(B_2)+f(B_3)+f(B_4)$$