4
$\begingroup$

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

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

$\endgroup$
9
  • $\begingroup$ Did you try Lagrange multipliers ? $\endgroup$ – Belgi Sep 14 '12 at 16:39
  • $\begingroup$ It's equivalent to $(\sum a^6)^3\ge3(\sum a^7b^2)^2$, which seems hard to prove. $\endgroup$ – Yai0Phah Sep 14 '12 at 17:18
  • $\begingroup$ @FrankScience I will thank if you tell me how you obtain this equivalence . Thanks :) $\endgroup$ – Iuli Sep 14 '12 at 17:21
  • $\begingroup$ @Iuli: are $a, b, c$ positive real numbers, or are they non-negative real numbers? It makes a difference. $\endgroup$ – Rod Carvalho Sep 14 '12 at 17:25
  • $\begingroup$ @Iuli As $a,b,c>0$, it's equivalent to $(\sum a^7b^2)^2\le9=(\sum a^6)^3/3$. $\endgroup$ – Yai0Phah Sep 14 '12 at 17:27
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $(0)$

$\endgroup$

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.