Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$

This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find this inequality have solve it,maybe it seem can use integral to solve it?

my attempts:

I took $p=3a+2b,$ $2a+2b+c$. $k=3$ and I wanted to use this integral: $$\dfrac{1}{p^k}=\dfrac{1}{\Gamma(k)}\int_{0}^{+\infty}t^{k-1}e^{-pt}dt$$ but I don't see how it helps.

  • 1
    $\begingroup$ I think it was a typo in the formulation. This inequality is true, but has a very ugly solution. The inequality with nice solution is the following. $\sum\limits_{cyc}\frac{a^3b}{(2a+3b)^3}\geq\sum\limits_{cyc}\frac{a^2bc}{(2a+2b+c)^3}.$ $\endgroup$ – Michael Rozenberg Apr 3 '19 at 4:59
  • $\begingroup$ can you post your nice solution with $\sum_{cyc}\dfrac{a^3b}{(2a+3b)^3}\ge\sum_{cyc}\dfrac{a^2bc}{(2a+2b+c)^3}$?Thanks $\endgroup$ – function sug Apr 3 '19 at 5:00
  • 1
    $\begingroup$ @MichaelRozenberg,I have add mu attempts,and take $p=3a+2b$,$2a+2b+c$.$k=3$,maybe can help? $\endgroup$ – function sug Apr 3 '19 at 5:41
  • 1
    $\begingroup$ This inequality is obviously true by BW, but this solution is not by hand. $\endgroup$ – Michael Rozenberg Apr 10 '19 at 15:10
  • 1
    $\begingroup$ Also, this inequality is obviously true after full expanding by AM-GM. But for the proof we need to use a computer again. $\endgroup$ – Michael Rozenberg Apr 10 '19 at 20:43

The proof of my inequality.

let $a$, $b$ and $c$ be positive numbers. Prove that: $$\tfrac{a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+2c)^3}+\tfrac{c^3a}{(2c+3a)^3}\geq\tfrac{a^2bc}{(2a+2b+c)^3}+\tfrac{b^2ca}{(2b+2c+a)^3}+\tfrac{c^2ab}{(2c+2a+b)^3}.$$

Indeed, by Holder and AM-GM we obtain: $$\sum_{cyc}\tfrac{a^3b}{(2a+3b)^3}=\sum_{cyc}\tfrac{\left(4(2a+3b)+(2b+3c)+2(2c+3a)\right)^3\left(\tfrac{4a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+3c)^3}+\tfrac{2c^3a}{(2c+3a)^3}\right)}{2401(2a+2b+c)^3}\geq$$ $$\geq\sum_{cyc}\frac{\left(4\sqrt[4]{a^3b}+\sqrt[4]{b^3c}+2\sqrt[4]{c^3a}\right)^4}{2401(2a+2b+c)^3}\geq\sum_{cyc}\frac{\left(7\sqrt[28]{a^{12+2}b^{4+3}c^{1+6}}\right)^4}{2401(2a+2b+c)^3}=\sum_{cyc}\frac{a^2bc}{(2a+2b+c)^3}.$$


Using binomial inequality for $$|c-a|<2a+2b+c,$$ one can get $$\dfrac1{(3a+2b)^3}=\dfrac1{(2a+2b+c)^3}\left(1-\dfrac{c-a}{2a+2b+c}\right)^{-3} \ge \dfrac1{(2a+2b+c)^3}\left(1+3\dfrac{c-a}{2a+2b+c}\right),$$ $$\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3} \ge \dfrac{a^2b}{(2a+2b+c)^3}\left(a-c-3a\dfrac{c-a}{2a+2b+c}\right),\tag1$$ The issue inequality can be presented in the form of $$S\ge 0\tag2,$$ where $$S = \sum\limits_\bigcirc \left(\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3}\right) \ge \sum\limits_\bigcirc\dfrac{a^2b(a-c)}{(2a+2b+c)^3} +3\sum\limits_\bigcirc\dfrac{a^3b(a-c)}{(2a+2b+c)^4}.$$

In according with the rearrangement inequality for the productions of $$a\cdot a\cdot a\cdot \dfrac b{2a+2b+c} + b\cdot b\cdot b\cdot \dfrac c{2b+2c+a} + c\cdot c\cdot c\cdot \dfrac a{2c+2a+b}$$ and $$a\cdot a\cdot c\cdot \dfrac b{2a+2b+c} + b\cdot b\cdot a\cdot \dfrac c{2b+2c+a} + c\cdot c\cdot b\cdot \dfrac a{2c+2a+b}$$ between $a$ and $c$ (where the ratio can be placed arbitrary), one can write $$\sum\limits_\bigcirc\dfrac{a^3b}{(2a+2b+c)^3} \ge \sum\limits_\bigcirc\dfrac{a^2bc}{(2a+2b+c)^3}$$ and similarly
$$\sum\limits_\bigcirc\dfrac{a^4b}{(2a+2b+c)^4} \ge \sum\limits_\bigcirc\dfrac{a^3bc}{(2a+2b+c)^4},$$
so $(1)$ is valid.


  • $\begingroup$ To which numbers exactly is the rearrangement inequality applied? $\endgroup$ – Martin R Apr 17 '19 at 12:03
  • $\begingroup$ @MartinR To subsequences $\{a,a\}$ and $\{c.c\}.$ $\endgroup$ – Yuri Negometyanov Apr 17 '19 at 12:08
  • $\begingroup$ Well, perhaps I am too dumb to see it. The rearrangement inequality states that $x_1 y_1 + x_2 y_2 + x_3 y_3$ is maximal if $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ are ordered in the same way (both ascending or both descending). What are $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ in your case so that the final two inequalities are obtained? $\endgroup$ – Martin R Apr 17 '19 at 12:22
  • $\begingroup$ @MartinR I think that the most clear way is to correspond the debominators with $b.$ $\endgroup$ – Yuri Negometyanov Apr 17 '19 at 12:31
  • 1
    $\begingroup$ @Yuri Negometyanov Can you write sequences, for which you used Rearrangement. Your sequences $(a,a)$ and $(c,c)$ give $a^2+c^2\geq2ac$, which is true, but it's nothing here. Actually, components of sequences should be ordered, otherwise Rearrangement does not work. $\endgroup$ – Michael Rozenberg Apr 17 '19 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.