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Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers.

First, I tried to simplify the proof statement but I got an even more compliicated: $$a^4b^4+b^4c^4+a^4c^4> a^2b^3c^3+b^2c^3a^3+a^3b^3c^2$$

Then I used Power mean inequality on $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac {1}{c}$ but that wasn't useful here.

Finally, I tried to solve it using AM-HM inequality but couldn't.

What would be an efficient method to solve this problem? Please provide only a hint and not the entire solution since I wish to solve it myself.

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    $\begingroup$ $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3} =abc\left(\dfrac{1}{c^4}+\dfrac{1}{a^4}+\dfrac{1}{b^4}\right) $ $\endgroup$ – marty cohen Oct 15 '17 at 19:58
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    $\begingroup$ I think it's duplicate $\endgroup$ – user334639 Oct 15 '17 at 20:06
  • $\begingroup$ @user334639 I think every math problem it's a duplicate because it's a math problem. $\endgroup$ – Michael Rozenberg Oct 16 '17 at 0:57
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    $\begingroup$ @Did what's a PSQ ? $\endgroup$ – Abcd Oct 16 '17 at 8:49
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    $\begingroup$ Sorry, folks. I cannot find the "duplicate". Probably it was a déjà vu of a similar inequality... $\endgroup$ – user334639 Oct 16 '17 at 12:00
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AM-GM helps! $$\sum_{cyc}\frac{ab}{c^3}=\frac{1}{4}\sum_{cyc}\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)\geq\frac{1}{4}\sum_{cyc}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}\right)=\sum_{cyc}\frac{1}{c}.$$ Done!

Without $cyc$ we can write the solution so: $$\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}=$$ $$=\frac{1}{4}\left(\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{2bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{2ca}{b^3}\right)\right)\geq$$ $$\geq\frac{1}{4}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{bc}{a^3}\right)^2\cdot\frac{ab}{c^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{ca}{b^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ab}{c^3}}\right)=$$ $$=\frac{1}{c}+\frac{1}{a}+\frac{1}{b}.$$

The same trick gives also a proof by Holder: $$\sum_{cyc}\frac{ab}{c^3}=\sqrt[4]{\left(\sum_{cyc}\frac{ab}{c^3}\right)^2\sum_{cyc}\frac{bc}{a^3}\sum_{cyc}\frac{ca}{b^3}}\geq\sum_{cyc}\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}=\sum_{cyc}\frac{1}{c}.$$ Turned out even a bit of shorter.

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    $\begingroup$ What does $\sum_{cyc}$ mean??? $\endgroup$ – Mr Pie Oct 16 '17 at 1:17
  • $\begingroup$ @user477343 $a\rightarrow b\rightarrow c\rightarrow a$. For example, $\sum\limits_{cyc}a=a+b+c$ or $\sum\limits_{cyc}a^2b=a^2b+b^2c+c^2a$ or $\sum\limits_{cyc}\frac{ab}{c^3}=\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}.$ $\endgroup$ – Michael Rozenberg Oct 16 '17 at 1:21
  • $\begingroup$ Thank you for that. You have now technically answered two questions :) $\endgroup$ – Mr Pie Oct 16 '17 at 1:24
  • $\begingroup$ why is there a $\dfrac{1}{4}$ in the first step? $\endgroup$ – Abcd Oct 16 '17 at 4:29
  • $\begingroup$ @Abcd It's designation only. See please my previous comment. I am ready to explain. $\endgroup$ – Michael Rozenberg Oct 16 '17 at 4:30
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Use the rearrangement inequality. Assume without loss of generality $a\geq b\geq c>0$. Then we have $ab\geq ac\geq bc\,$ and $\,1/c^3\geq 1/b^3\geq 1/a^3$. Therefore the sorted sum-product

$$\frac{ab}{c^3}+\frac{ac}{b^3}+\frac{bc}{a^3}\geq\frac{bc}{c^3}+\frac{ab}{b^3}+\frac{ac}{a^3}=\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}{a^2}$$

is greater than equal to a shuffled sum-product, which is greater than equal to the reversed sum-product

$$\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}{a^2}\geq\frac{c}{c^2}+\frac{b}{b^2}+\frac{a}{a^2}=\frac{1}{c}+\frac{1}{b}+\frac{1}{a}.$$

Equality is obtained if and only if $\,a=b=c$. If $a,b,c$ are different (not necessarily all different), the greater than "$>$" sign holds.

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  • $\begingroup$ Maybe I don't know the correct name. It says for positive numbers $0<a_1\leq\cdots\leq a_n$ and $0<b_1\leq\cdots\leq b_n$, you have $\sum_ia_ib_i\geq \sum_ia_ib_{p_i}\geq\sum_i a_ib_{n+1-i}$ for an arbitrary permutation $p_i$ of the indices $1\cdots n$. $\endgroup$ – Zhuoran He Oct 15 '17 at 20:16
  • $\begingroup$ @Abcd I found the correct name: en.wikipedia.org/wiki/Rearrangement_inequality. And in fact it applies to all real numbers. $\endgroup$ – Zhuoran He Oct 15 '17 at 20:33
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\begin{eqnarray*} a^4(b^2-c^2)^2+b^4(a^2-c^2)^2+c^4(b^2-a^2)^2+a^2b^2c^2((a-b)^2+(c-b)^2+(a-c)^2) \geq 0. \end{eqnarray*} Rearrange to \begin{eqnarray*} a^4b^4+b^4c^4+c^4a^4 \geq a^2b^2c^2(ab+bc+ca). \end{eqnarray*} Now divide by $a^3b^3c^3$ and we have \begin{eqnarray*} \frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{ b^3} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \end{eqnarray*}

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  • $\begingroup$ How did you get the first statement? $\endgroup$ – Abcd Oct 15 '17 at 20:14
  • $\begingroup$ To get the second formula is just algebra ... I then drew a triangular grid ($9$ points on each edge) and plotted the terms, I then started looking for exact squares that would contain the negative terms. $\endgroup$ – Donald Splutterwit Oct 15 '17 at 20:18
  • $\begingroup$ I call this method "brute force" ... just reduce it to the sum of positive terms. If you wait a little while ... folks will derive your inequality using clever results like CS or AM-HM etc... $\endgroup$ – Donald Splutterwit Oct 15 '17 at 20:21

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