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If $a,b,c$ are positive real numbers, prove $$\sum \limits_{cyc} \frac{1+b^2+c^4}{a+b^2+c^3}\geq 3$$ Additional info: We should only use Cauchy (preferred to used at least once and more than AM-GM) and AM-GM. We are not allowed to use induction.

Things I have done so far: My idea is about separating question into $3$ inequality and proving them one by one and then sum all of them to prove question inequality.So$$\sum \limits_{cyc} \frac{1}{a+b^2+c^3} \geq 1$$$$\sum \limits_{cyc} \frac{b^2}{a+b^2+c^3} \geq 1$$ $$\sum \limits_{cyc} \frac{c^4}{a+b^2+c^3} \geq 1$$

UPDATE

As Macavity pointed in comments, the separating idea is not useful because first inequality does not hold.So any hint for starting step is appreciated.

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  • $\begingroup$ Note that the first of your three split equalities will not hold for all reals. Just take $a,b,c$ Large enough to see that. $\endgroup$ – Macavity Jul 31 '14 at 13:36
  • $\begingroup$ @Macavity,thanks for pointing out.what about question inequality?,does it have counter example? $\endgroup$ – user2838619 Jul 31 '14 at 13:38
  • $\begingroup$ The problem statement looks good, can't find any counter eg readily. $\endgroup$ – Macavity Jul 31 '14 at 13:39
  • $\begingroup$ @Macavity,what does SOS stand for? $\endgroup$ – user2838619 Jul 31 '14 at 16:44
  • $\begingroup$ In its basic form, expressing some expression as a Sum Of Squares, thereby certifying it is $\ge 0$. $\endgroup$ – Macavity Jul 31 '14 at 16:47
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Using Cauchy-Schwarz, note $$LHS = \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a+b^2+c^3)^2} \ge \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a^2+b^2+c^2)(1+b^2+c^4)} = \sum_{cyc} \frac{a+b^2+c^3}{a^2+b^2+c^2}$$

As the denominators of the cyclic sum are now the same, it only remains to show that $$\sum_{cyc} (a + b^2+c^3) \ge 3(a^2+b^2+c^2) \iff \sum_{cyc} a + \sum_{cyc} a^3 \ge 2\sum_{cyc} a^2$$ which follows from AM-GM as $a+a^3 \ge 2a^2$.

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    $\begingroup$ I am speechless,awesome proof. $\endgroup$ – user2838619 Aug 1 '14 at 11:58

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