Prove that $a^2bc + ab^2c + abc^2 \le a^3b+ac^3+b^3c$

Question

Note: $a,b$ and $c$ are positive real numbers.

I tried to use excel and I believe that, after going through a bunch of numbers, this preposition is true. However, I do not know how to prove it mathematically. Can someone help me to prove this question.

For what numbers do you think it is true? I assume you tried only with positive numbers, as with $a=0$, $b=1$ and $c=-1$ the inequality does not hold. — TheDoyduro, Jun 23 '20 at 12:04
Oh no, I forgot to add in the extra conditions. Let me edit that. — MJ2020, Jun 23 '20 at 12:06
Familiar with Muirhead's inequality? Above is a partial case for $a_1=2,a_2=1,a_3=1$ and $b_1=3,b_2=1,b_3=0$ — Alexey Burdin, Jun 23 '20 at 12:31
Does this answer your question? prove the inequality for positive a, b, c — Martin R, Jun 23 '20 at 12:50

score 4 · Answer 1 · answered Jun 23 '20 at 12:33

You have to show

$$abc(a+b+c)\leq a^3b+ac^3+b^3c$$

or (since $a,b,c > 0)$

$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$

Now, Cauchy-Schwarz inequality helps as follows

$$(a+b+c)^2 = \left(\sqrt c\frac a{\sqrt c}+\sqrt a \frac b{\sqrt a }+\sqrt b \frac c{\sqrt b } \right)^2\stackrel{C.-S.}{\leq} (c+a+b)\cdot \left(\frac{a^2}{c}+ \frac{b^2}{a} + \frac{c^2}{b} \right)$$

Hence,

$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$

score 1 · Answer 2 · answered Jun 23 '20 at 12:44

Since $a,b,c$ are positive, dividing both sides by $abc$ we get the equivalent inequality $$a+b+c\leq\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\tag{1}$$

Without loss of generality let $$a\leq b\leq c$$ Then $$a^2\leq b^2\leq c^2$$ and $$\frac{1}{c}\leq\frac{1}{b}\leq\frac{1}{a}$$ Hence $(1)$ is true by rearrangement inequality. See here for details https://en.m.wikipedia.org/wiki/Rearrangement_inequality

Prove that $a^2bc + ab^2c + abc^2 \le a^3b+ac^3+b^3c$

2 Answers2