Prove $\sum \limits_{cyc} \frac{1+b^2+c^4}{a^+b^2+c^3}\geq 3$

Question

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.

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. — Macavity, Jul 31 '14 at 13:36
@Macavity,thanks for pointing out.what about question inequality?,does it have counter example? — user2838619, Jul 31 '14 at 13:38
The problem statement looks good, can't find any counter eg readily. — Macavity, Jul 31 '14 at 13:39
In its basic form, expressing some expression as a Sum Of Squares, thereby certifying it is $\ge 0$. — Macavity, Jul 31 '14 at 16:47

score 8 · Accepted Answer · answered Aug 01 '14 at 09:37

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 $a+a^3 \ge 2a^2$.

I am speechless,awesome proof. – user2838619 Aug 01 '14 at 11:58 — user2838619, Aug 01 '14 at 11:58

Prove $\sum \limits_{cyc} \frac{1+b^2+c^4}{a^+b^2+c^3}\geq 3$

1 Answers1

Linked