For $a,b,c>0$ prove that $abc(a+b+c) \le a^3 b + b^3 c + c^3 a$

Question

if $a,b,c > 0$ then prove that $abc(a+b+c) \le a^3 b + b^3 c + c^3 a$

My attempt

The hint given for this question was to use the Cauchy-schwarz inequality. But if you look at the expression given, the L.H.S is symmetric but the R.H.S is not. If I understand the inequality(Cauchy-Schwarz) correctly, both the sides of the inequality should be symmetric.

So can we use the Cauchy-Schwarz to prove this?

I also tried applying AM > GM but got stuck.

Any help would be appreciated.

Answer 1

We can write this inequality as $$F=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b} \ge (a+b+c)~~~(1)$$ Using Cauch-Schwarts ineq. : $$\frac{A^2}{X}+\frac{B^2}{Y}+\frac{C^2}{Z} \ge \frac{(A+B+C)^2}{X+Y+Z}, X,Y,Z>0~~~~(2)$$ Using (2) it follows that $$F\ge \frac{(a+b+c)^2}{(a+b+c)}=a+b+c.$$

Answer 2

We need to show $$abc(a+b+c)\leq a^3b+b^3c+c^3a$$for all $a,b,c\in\mathbb R^+$. The proof uses rearrangement inequality by considering two case $a\geq b\geq c$ and $a\geq b\geq c$. I will show how this works for $a\geq b\geq c$ (the proof of second case follows the same way).

Note that $a\geq b\geq c\implies ab\geq ac\geq bc, a^2\geq b^2\geq c^2$. Thus, by rearrangement inequality, we get $$(a^2)(ab)+(b^2)(bc)+(c^2)(ca)\geq (a^2)(bc)+(b^2)(ac)+(c^2)(ab)=abc(a+b+c)$$

PS: This can also be proved using $T_2$'s lemma.

Answer 3

By AM-GM: $$\sum_{cyc}\left(a^3b-a^2bc\right)=\sum_{cyc}(a^3b+c^2ab-2a^2bc)\geq\sum_{cyc}\left(2\sqrt{a^3b\cdot c^2ab}-2a^2bc\right)=0.$$

Answer 4

AM-GM works, you just have to rearrange the terms:

$ 4 a^3 b + 1 b^3 c + 2 c^3 a \geq 7 \sqrt[7]{a^{14}b^7c^7} = 7 a^2 bc$.

Add the 3 cyclic inequalities, and we conclude that

$a^3 b + b^3 c + c^3 a \geq abc(a+b+c)$

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Note: The inequality Muirhead's inequality, which can be proved via similar AM-GM bunching.