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Let $a,b,c$ be positive real numbers such that $a^6+b^6+c^6=3$. Prove that

$$a^7b^2+b^7c^2+c^7a^2 \leq 3 .$$

Community
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Iuli
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  • Did you try Lagrange multipliers ? – Belgi Sep 14 '12 at 16:39
  • It's equivalent to $(\sum a^6)^3\ge3(\sum a^7b^2)^2$, which seems hard to prove. – Yai0Phah Sep 14 '12 at 17:18
  • @FrankScience I will thank if you tell me how you obtain this equivalence . Thanks :) – Iuli Sep 14 '12 at 17:21
  • @Iuli: are $a, b, c$ positive real numbers, or are they non-negative real numbers? It makes a difference. – Rod Carvalho Sep 14 '12 at 17:25
  • @Iuli As $a,b,c>0$, it's equivalent to $(\sum a^7b^2)^2\le9=(\sum a^6)^3/3$. – Yai0Phah Sep 14 '12 at 17:27
  • @FrankScience I will appreciate if you give a more detailed answer. Thanks:) – Iuli Sep 14 '12 at 17:30
  • Just type the following into WolframAlpha: optimize a^7 b^2 + b^7 c^2 + c^7 a^2 subject to a^6 + b^6 + c^6 = 3]. I'm sure that it uses Lagrange multipliers. If all you need is verification, then you're done. If this is homework, then you should try a bit harder! - Hmm... I wonder how to put a link into a comment. – Mark McClure Sep 14 '12 at 19:12
  • Why do you think that this is true in the first place? The best questions on this site explain the motivation behind the question, how it was discovered, and they say what you have tried already. This question simply commands others to prove something, which can appear to be impolite, and at the same time has no context at all. For this reason, I have voted -1. – Carl Mummert Sep 16 '12 at 02:37
  • @CarlMummert I admire your justice. Please excuse me, you are so perfect. I will send you all the questions from all the users which didnt't try anything to their question. If you had looked to my questions you would have seen that I use to write what I have tried, but where I made only calculations without to obtain something I didn't write - if I wrote $n$ pages, is it normally to write them on the site? P.S. I posted few questions without saying what I have tried. I am waiting for your (-1) vote. – Iuli Sep 16 '12 at 06:40

2 Answers2

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Using AM-GM inequality, we get $$ 3 = \frac {(a^6 + b^6 + c^6)^2} 3 = \sum_{cyc} a^6 \frac {a^6 + 2b^6} 3 \geq \sum_{cyc} a^6\sqrt[3]{a^6 b^{12}} = a^8b^4 + b^8c^4 + c^8a^4 $$ Now, by means of Cauchy–Schwarz inequality we complete the proof $$ a^3 \cdot a^4 b^2 + b^3 \cdot b^4 c^2 + c^3 \cdot c^4 a^2 \leq \sqrt{a^6 + b^6 + c^6}\sqrt{a^8 b^4 + b^8 c^4 + c^8 a^4} \leq 3 $$

AlbertH
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Frank Science has showed that the inequality is equivalent to $$ 3(\sum a^7b^2)^2 \leqslant (\sum a^6)^3\tag{0}\\ $$

Use AM-GM for $a^{12}b^{6}$, $a^{12}b^{6}$ and $a^{18}$ we have the following:

$ 3a^{14}b^4 \leqslant 2a^{12}b^6+a^{18}\\ $

Combine with 2 other similar inequalities we have

$$3(\sum a^{14}b^4) \leqslant \sum a^{18}+2(\sum a^{12}b^6)\tag{1}$$

Use AM-GM again this time we have the following:

$ 6a^7b^9c^2 \leqslant 2(abc)^6+a^{12}b^6+3a^6b^{12}\\ $

Combine with 2 other similar inequalities we have

$$6(\sum a^7b^9c^2) \leqslant 6(abc)^6+\sum a^{12}b^6+3(\sum a^6b^{12}) \tag{2}$$

Summing $(1)$ and $(2)$ we have $(0)$

Iuli
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Minh Hoang
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