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Prove that $a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3$

SOURCE : "A Brief Introduction to Olympiad Inequalities" by Evan Chen

It was one of the practice problems. Equality case is easy. I tried AM-GM and Muirhead, but could not seem to find a suitable proof for the inequality.

Any hint would be really helpful.

Thanks in advance !

^__^

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    are $$a,b,c$$ positive real? – 2017-01-29
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    it can be solved with BW – 2017-01-29
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    @Dr Sonnard Graubner: I think that positive real because if $a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3$ for positive then the inequality is changed for negative, i. e. $a^7+b^7+c^7\le a^4b^3+b^4c^3+c^4a^3$ for the same values but with distinct sign. – 2017-01-29
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    I guess [*rearrangement inequality*](https://en.wikipedia.org/wiki/Rearrangement_inequality) may be helpful as well. – 2017-01-29
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    @Dr.SonnhardGraubner Nothing has been mentioned ... you can check out the link if you want. Also, what do you mean by "BW" ?? – 2017-01-29

3 Answers 3

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Using $\text{AM-GM}$ this can be solved. Note that by $\text{AM-GM}$ $$4a^7+3b^7=a^7+a^7+a^7+a^7+b^7+b^7+b^7 \ge 7a^4b^3$$$$4b^7+3c^7=b^7+b^7+b^7+b^7+c^7+c^7+c^7 \ge 7b^4c^3$$$$4c^7+3a^7=c^7+c^7+c^7+c^7+a^7+a^7+a^7 \ge 7c^4a^3$$ Adding the three and dividing by seven, we have $$a^7+b^7+c^7 \ge a^4b^3+b^4c^3+c^4a^3$$

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    OH MY GOD !!!! Such a beautiful solution .... :) +1. – 2017-01-29
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    Right. Beautiful really. – 2017-01-29
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It's a direct result of the rearrangement inequality, i.e.

$$ S_n = \sum_{i = 1}^n a_ib_i$$ is maximal if sequences $ \{a_i\}$ and $ \{b_i\}$ are similarly sorted; and minimal if sequences $ \{a_i\}$ and $ \{b_i\}$ are oppositely sorted.

Given $\{a,b,c\}$, $\{a^3,b^3,c^3\}$ and $\{a^4,b^4,c^4\}$ are similarly sorted, we can rewrite $a^7+b^7+c^7=a^4a^3+b^4b^3+c^4c^3$ which has the maximal value.

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    You can not assume that $a\geq b\geq c$ because our inequality is cyclic an not symmetric. But indeed, $(a^4,b^4,c^4)$ and $(a^3,b^3,c^3)$ are the same ordered and Rearrangement works. – 2017-01-29
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    What I meant is that we can equally assume $a\geq b\geq c$ or $a\geq c\geq b$ or $b\geq c \geq a$ etc. and $\{a^4,b^4,c^4\}$ and $\{a^3,b^3,c^3\}$ will be equally sorted. – 2017-01-29
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    Just your words "Without loss of generality we can assume that a≥b≥c" say another. OK. I understood you. – 2017-01-29
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    Ok, I changed the wording, hopefully it's clearer now ... – 2017-01-29
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There is the following general technique. $$\sum_{cyc}(a^7-a^4b^3)=\sum_{cyc}a^4(a-b)(a^2+ab+b^2)=$$ $$=\sum_{cyc}\left(a^4(a-b)(a^2+ab+b^2)-\frac{3}{7}(a^7-b^7)\right)=$$ $$=\frac{1}{7}\sum_{cyc}(a-b)^2(4a^5+8a^4b+12a^3b^2+9a^2b^3+6ab^4+3b^5)\geq0$$