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

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

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

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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 $(0)$

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