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Let $a,b,c>0$. Show that $$a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3\geq{\frac{8abc(a+b+c)^2}{3}}.$$

I have no idea. Please give a hint. Thanks!

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closed as off-topic by user99914, jvdhooft, José Carlos Santos, Chris Godsil, Taroccoesbrocco Jun 16 '18 at 1:39

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Proof

By expanding and rewriting, the inequality we want to prove is equivalent to

$3(a^5+b^5+c^5-a^2b^2c-a^2bc^2-ab^2c^2)+8(a^4b+b^4c+c^4a-a^3bc-ab^3c-abc^3)+(a^4b+b^4c+c^4a-a^2b^2c-a^2bc^2-ab^2c^2)+9(a^3b^2+b^3c^2+c^3a^2-a^2b^2c-a^2bc^2-ab^2c^2)+3(a^3c^2+b^3a^2+c^3b^2-a^2b^2c-a^2bc^2-ab^2c^2)\geq 0.$ $\tag0$

Step 1 Show that

$a^5+b^5+c^5 \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(bc+ca+ab)\left(\frac{a^4}{bc}+\frac{b^4}{ca}+\frac{c^4}{ab}\right)\geq (a^2+b^2+c^2)^2 \geq (ab+bc+ca)^2.\tag1$$ Thus,$$\frac{a^4}{bc}+\frac{b^4}{ca}+\frac{c^4}{ab} \geq ab+bc+ca.\tag2$$ As a result,$$a^5+b^5+c^5 \geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag3$$

Step 2 Show that

$a^4b+b^4c+c^4a \geq a^3bc+ab^3c+abc^3$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(ca+ab+bc)\left(\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b}\right)\geq (a^2+b^2+c^2)^2 \geq (ab+bc+ca)(a^2+b^2+c^2).\tag4$$ Thus,$$\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b} \geq a^2+b^2+c^2.\tag5$$ As a result,$$a^4b+b^4c+c^4a \geq abc(a^2+b^2+c^2)=a^3bc+ab^3c+abc^3.\tag6$$

Step 3 Show that

$a^4b+b^4c+c^4a \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c >0$.

From $(6)$, we obtain$$a^4b+b^4c+c^4a \geq abc(a^2+b^2+c^2)\geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag7$$

Step 4 Show that

$a^3b^2+b^3c^2+c^3a^2 \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(bc+ca+ab)\left(\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\right) \geq (ab+bc+ca)^2.\tag8$$ Thus,$$\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b} \geq ab+bc+ca.\tag9$$ As a result,$$a^3b^2+b^3c^2+c^3a^2\geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag{10}$$

Step 5 Show that

$a^3c^2+b^3a^2+c^3b^2 \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(bc+ca+ab)\left(\frac{a^2c}{b}+\frac{b^2a}{c}+\frac{c^2b}{a}\right) \geq (ab+bc+ca)^2.\tag{11}$$ Thus,$$\frac{a^2c}{b}+\frac{b^2a}{c}+\frac{c^2b}{a} \geq ab+bc+ca.\tag{12}$$ As a result,$$a^3c^2+b^3a^2+c^3b^2\geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag{13}$$

According to the five steps above, we can see that $(0)$ holds for $a,b,c>0.$

Please correct me if I'm wrong!

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  • $\begingroup$ @Michael Rozenberg Sir, please review my proof,THX. $\endgroup$ – mengdie1982 May 31 '18 at 10:03
  • $\begingroup$ I think your proof is right. $\endgroup$ – Michael Rozenberg May 31 '18 at 10:20
  • $\begingroup$ @mengdie1982 Hi. If you don't mind, can I please ask some questions about your solution? 1) How did you know that you should expand, rather than try some tricks on the original inequality, maybe with some mean inequalities? 2) After expanding, how did you know to break it up like you did? $\endgroup$ – Ovi Jun 2 '18 at 1:37
  • $\begingroup$ @Ovi Just try and try... But I also want to see another proof without expanding it entirely. $\endgroup$ – mengdie1982 Jun 2 '18 at 8:00
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The hint:

After full expanding we need to prove that $$\sum_{cyc}(3a^5+9a^4b+9a^3b^2+3a^3c^2-8a^3bc-16a^2b^2c)\geq0.$$ Now, prove by Rearrangement that $$\sum_{cyc}a^4b\geq\sum_{cyc}a^3bc$$ and the rest it's Rearrangement again or AM-GM.

Good luck!

Also, we can use Holder and C-S: $$\sum_{cyc}a^2(a+b)^3=\sum_{cyc}\frac{a^3(a+b)^3}{a}\geq\frac{\left(\sum\limits_{cyc}a(a+b)\right)^3}{3(a+b+c)}\geq$$ $$\geq\frac{\left(\frac{6}{9}\sum\limits_{cyc}(a^2+2ab)\right)^3}{3(a+b+c)}=\frac{8(a+b+c)^5}{81}\geq\frac{81abc(a+b+c)^2}{3}.$$

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  • $\begingroup$ Hello! Do you think there is some trick to solve this question without expanding it out? $\endgroup$ – Ovi Jun 2 '18 at 1:38
  • $\begingroup$ @Ovi I added something. See now. $\endgroup$ – Michael Rozenberg Jun 2 '18 at 13:05
  • $\begingroup$ Ah very nice! ${}$ $\endgroup$ – Ovi Jun 2 '18 at 14:26
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Another Proof

By Cauchy's Inequality, we have \begin{align*} &[(a+b)+(b+c)+(c+a)][a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3]\\ \geq &[a(a+b)^2+b(b+c)^2+c(c+a)^2]^2,\end{align*}and \begin{align*} &(a+b+c)[a(a+b)^2+b(b+c)^2+c(c+a)^2]\\ \geq &[a(a+b)+b(b+c)+c(c+a)]^2.\end{align*}Besides, notice that $$a(a+b)+b(b+c)+c(c+a)\geq \frac{2}{3}(a+b+c)^2,$$ and $$(a+b+c)^3 \geq 27abc.$$ It follows that \begin{align*} a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3 &\geq \frac{[a(a+b)^2+b(b+c)^2+c(c+a)^2]^2}{2(a+b+c)}\\ &\geq \frac{[a(a+b)+b(b+c)+c(c+a)]^4}{2(a+b+c)^3}\\ &\geq \frac{\left[\dfrac{2}{3}(a+b+c)^2\right]^4}{2(a+b+c)^3}\\&=\frac{8}{81}(a+b+c)^3(a+b+c)^2\\&\geq \frac{8}{81}\cdot 27abc \cdot (a+b+c)^2\\&=\frac{8}{3}abc(a+b+c)^2. \end{align*}

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  • $\begingroup$ @Ovi Am I right? $\endgroup$ – mengdie1982 Jun 2 '18 at 12:04

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