For positive reals $a$, $b$ and $c$, prove that $a^3b + b^3c + c^3a \ge abc(a+b+c)$.

Question

For positive reals $a$, $b$ and $c$, prove that $a^3b + b^3c + c^3a \ge abc(a+b+c)$.

Well Chebysev's inequality looks natural here though it doesn't seems that easy to find the right order of $(a^2b, b^2c, c^2a)$ like if W.L.O.G we assume $a \ge b \ge c$ then clearly $a^2b \ge b^2c$ but how to prove $b^2c \ge c^2a$ (to put it in the non-increasing order of sequence) ?

It's possible to have $a \geq b \geq c$ with $b^2 c > c^2 a$ or with $b^2 c < c^2 a$. — aschepler, Nov 14 '24 at 12:37
Yes exactly then @aschepler how chebyshev could be used here (since we can't assume $b^2c > c^2a$) ? — Ash_Blanc, Nov 14 '24 at 12:39
This question is similar to: If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) $, found using https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24a%5E3b%20%2B%20b%5E3c%20%2B%20c%5E3a%20%5Cge%20abc(a%2Bb%2Bc)%24&p=1 — Anne Bauval, Nov 14 '24 at 13:18
@AnneBauval well there are several answers but I wanted one valid one with chebyshev-inequality — Ash_Blanc, Nov 14 '24 at 13:23
Chebyshev does not help when there is only cyclicity and not total symmetry. In other words, you cannot assume $a\geq b\geq c.$ Even if you only assume the max or min and tried to do a case work, one of the cases will not yield to Chebyshev or rearrangement. — dezdichado, Nov 14 '24 at 14:16
You should be explicit that you want a solution by chebyshev. That's not stated in the question. — Calvin Lin, Nov 14 '24 at 16:35
@CalvinLin but I wasn't sure if it's solvable with Chebyshev or not — Ash_Blanc, Nov 14 '24 at 16:51
That's fine. Not all problems are solvable. You should ask for what you want, and not wait till other people provide a solution and then "complain" that it's not what you want. — Calvin Lin, Nov 14 '24 at 17:00
There is a also a solution using the rearrangement inequality in the suggested duplicate target: https://math.stackexchange.com/a/3704360/42969 — Martin R, Nov 14 '24 at 18:22

score 3 · Answer 1 · answered Nov 14 '24 at 13:20

We know that $$ (a-c)^2\geq 0 $$ $$ a^2+c^2\geq 2ac $$ $$ \frac{a^2}{c}+c\geq 2a $$ likewise, we get that $$ \frac{b^2}{a}+a\geq 2b $$ $$ \frac{c^2}{b}+b\geq 2c $$ We sum everything and cancel one $a+b+c$ and have that $$\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\geq a+b+c $$ then multiply by $abc$ on both sides $$a^3b+b^3c+c^3a\geq abc(a+b+c) $$

score 1 · Answer 2 · edited Nov 14 '24 at 18:29

1

It can be also proved by using AM-GM as follows. In cyclic sums we have $$\sum\left(4a^3b+b^3c+2c^3a\right)\ge\sum 7\sqrt[7]{\left(a^3b\right)^4\cdot \left(b^3c\right)\cdot\left(c^3a\right)^2}= 7\sum a^2bc.$$

edited Nov 14 '24 at 18:29

Anne Bauval

49,005

answered Nov 14 '24 at 18:16

Vic_Dis

629

For positive reals $a$, $b$ and $c$, prove that $a^3b + b^3c + c^3a \ge abc(a+b+c)$.

2 Answers2