Prove that $\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3$

Question

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 !

^__^

Answer 1

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

Answer 2

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.

Answer 3

There is the following general technique. $$\sum_{cyc}(a^7-a^4b^3)=\sum_{cyc}a^4(a-b)(a^2+ab+b^2)=$$ $$=\sum_{cyc}\left(a^4(a-b)(a^2+ab+b^2)-\frac{3}{7}(a^7-b^7)\right)=$$ $$=\frac{1}{7}\sum_{cyc}(a-b)^2(4a^5+8a^4b+12a^3b^2+9a^2b^3+6ab^4+3b^5)\geq0$$