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Prove inequality with $a,b,c >0$ $$\frac{a^4}{a^3+b^3} + \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}$$

I tried the inequality: $\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c$, but seem like it doesn't help

Xeing
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  • Without loss of generality you may assume that $a=1$ and thus reduce to a two-variables problem. – Eckhard Mar 25 '13 at 14:13
  • @Eckhard: How can you assume that? Generality is lost there, I suppose. – Inceptio Mar 25 '13 at 14:18
  • @Inceptio: you suppose wrong. For every $\lambda>0$, the inequality is satisfied for a triple $(a,b,c)$ if and only if is satisfied for any triple $(\lambda a,\lambda b,\lambda c)$, because both sides of the inequality scale linearly in $\lambda$. In particular, you may take $\lambda=a^{-1}$ to reduce to the case $a=1$. – Eckhard Mar 25 '13 at 14:23
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    To reduce it to 2 variables, put $b=k_1a$ and $c=k_2a$. You can even assume $k_1 \leq k_2$ – Gautam Shenoy Mar 25 '13 at 14:28
  • @Eckhard: Right. I supposed wrong. – Inceptio Mar 25 '13 at 14:57
  • Since it is not symmetric, I think you probably need to discuss the order of the variables. I got $\sum(a^7b^3+a^3b^7) +\sum a^6(b^4-c^4) - abc\sum(a^5b^2+a^2b^5)$, the only problem is the asymmetric term, if you can prove at any time one of these three should be right, $\sum a^7b^3\ge \sum a^6c^4$ or $\sum a^3b^7\ge \sum a^6c^4$ or $\sum a^6b^4 \ge\sum a^6c^4$. – Yimin Mar 25 '13 at 19:44

1 Answers1

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There is an extensive discussion of this inequality here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=23113

I like the sum of squares method the best:

\begin{align} & \sum_{cyc}{\frac{a^4}{a^3+b^3}}-\frac{1}{2}\sum_{cyc}{a} \\ & =\frac{(a-b)^2(b-c)^2(c-a)^2(ab+ac+bc)^2}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)} \\ & +\sum_{cyc}{\frac{c^2(b-c)^2(2c^3(a^3+b^3)+c^2a^2(a+b)^2+2ca^4b+a^4(a^2+b^2))}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)}} \\ & \geq 0 \end{align}

Ivan Loh
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