Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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. –  Frank Science 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 $(\sum a^7b^2)^2\le9=(\sum a^6)^3/3$. –  Frank Science Sep 14 '12 at 17:27

2 Answers 2

up vote 2 down vote accepted

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 $$ a^3 \cdot a^4 b^2 + b^3 \cdot b^4 c^2 + c^3 \cdot c^4 a^2 \leq \sqrt{a^6 + b^6 + c^6}\sqrt{a^8 b^4 + b^8 c^4 + c^8 a^4} \leq 3 $$

share|improve this answer

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 $(0)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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