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Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$): $$ a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2 $$

and

$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc $$

They seems somewhat similar, so I hope there'a an exploitable link between them. They fall easily under Muirhead, yet I cannot figure out how to prove them using the Rearrangement Inequality.

Any hints greatly appreciated.

5 Answers 5

4

by AM-GM we get $$\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2}{6}\geq \sqrt[6]{a^{12}b^{12}c^{12}}=a^2b^2c^2$$

  • 0
    I must say, I was in fact aware of this solution, yet the problem is to employ the rearrangement inequality – 2017-02-02
  • 1
    your second inequality can be solved analogously – 2017-02-02
  • 0
    these are examples for using AM-GM in my opinion – 2017-02-02
3

Without loss of generality, $a\le b\le c$. Then $ab\le ac\le bc$. Denote $x_1=x_2=ab, x_3=x_4=ac, x_5=x_6=bc$, and $y_1=y_2=a,y_3=y_4=b,y_5=y_6=c$. Then both $x_i$ and $y_i$ are monotonic, and $x_1y_6+...+x_6y_1=6abc$, so it is the least sum among all rearrangements. It is not hard to find a rearrangement to get left-hand side of the second inequality. The first is the same, just change all the letters by their squares.

3

$(a,b,c)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are opposite ordered.

Thus, by Rearrangement $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a\cdot\frac{1}{a}+b\cdot\frac{1}{b}+c\cdot\frac{1}{c}=3,$$ which gives $$a^2c+b^2a+c^2b\geq3abc$$ Similarly we'll get $$a^2b+b^2c+c^2a\geq3abc$$ and after summing we are done!

3

Recall that if we have $x_0\leq \ldots \leq x_n$ and $y_0\leq \ldots \leq y_n$ then the Rearrangement Inequality tells us that

$$\sum_{i=0}^n x_iy_{n-i} \leq \sum_{i=0}^n x_iy_{\sigma(i)} \leq \sum_{i=0}^n x_iy_i$$ for any permutation $\sigma \in \mathfrak{S}_{n+1}$.

Since the inequalities involved are symmetric with respect to any permutation of $a,b,c$ we can assume WLOG that $a \geq b \geq c$. Thus $ab \geq ac \geq bc$.

Then let

$$x_0 = x_1 = x_2 = bc \qquad x_3 = x_4 = x_5 = ac \qquad x_6 = x_7 = x_8 = ab$$ $$y_0 = y_1 = y_2 = c \qquad y_3 = y_4 = y_5 = b \qquad y_6 = y_7 = y_8 = a$$

Then

$$\sum_{i=0}^8 x_iy_{8-i}\leq x_0y_8 + x_1y_5 + x_2y_2 + x_3y_7 + x_4y_4 + x_5y_1 + x_6y_6 + x_7y_3 + x_8y_0$$

i.e.

$$9abc \leq abc + b^2c + bc^2 + a^2c + abc + ac^2 + a^2b + ab^2 + abc$$

and so

$$6abc \leq b^2c + bc^2 + a^2c + ac^2 + a^2b + ab^2$$

The other inequality is equivalent.

  • 0
    Right the question says to use the rearrangement Inequality. I did not know what this was. Darth Geek your answer explains it nicely & utilises it well, so your answer is best. I am still interested by everyone elses solutions. – 2017-02-02
2

\begin{eqnarray*} a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0 \end{eqnarray*} and the second inequality follows. Now substitute $a^2$ for $a$ etc and the first inequality follows.