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

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!

Would you please provide your proof, or at least the outline of your proof? — Empiricist, Jan 16 '16 at 08:33
@ S. W. Cheung it's $c=d+u$, $b=d+u+v$, $a=d+u+v+w$, where $u$, $v$ and $w$ are non-negatives and a lot of work. — Michael Rozenberg, Jan 16 '16 at 08:59
@MichaelRozenberg: It would be an improvement to add the details of your proof (or proof sketch) to the Question itself, not merely as alluded to in a Comment. After all you are asking for something to be compared (in "beauty") to your own approach, so it forms an integral part of the Question's context. — hardmath, Jan 16 '16 at 11:59
yao4015's deleted answer gave a counterexample for Roman83's answer: Let $x_1=4, x_2=3,x_3=3,x_4=1$, then $A=(12,11,6,4), B=(12,9,9,3)$. But $A_1+A_2+A_3=29, B_1+B_2+B_3=30$, so $A \nsucc B$ . — River Li, Mar 06 '24 at 07:38

score 4 · Answer 1 · answered Apr 27 '16 at 17:43

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:

$$f(A_1)+f(A_2)+f(A_3)+f(A_4)\ge f(B_1)+f(B_2)+f(B_3)+f(B_4)$$

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