Note that $\frac{a^2 + 2}{2} = \frac{(a^2 - a + 1) + (a + 1)}{2} \geq \sqrt{(a^2 - a + 1)(a + 1)} = \sqrt{a^3 + 1}$, with equality when $a=2$. Hence it suffices to prove \[\frac{a^2}{(a^2 + 2)(b^2 + 2)} + \frac{b^2}{(b^2 + 2)(c^2 + 2)} + \frac{c^2}{(c^2 + 2)(a^2 + 2)} \geq \frac{1}{3}\] and this is easily verified.

Sorry. you said that the last inequality is easily verified. How can we prove it ?

erdos wrote: Sorry. you said that the last inequality is easily verified. How can we prove it ? Once we get through the first step, the rest is trivial. \[ \frac{a^2}{(a^2 + 2)(b^2 + 2)} + \frac{b^2}{(b^2 + 2)(c^2 + 2)} + \frac{c^2}{(c^2 + 2)(a^2 + 2)} \geq \frac{1}{3} \] The above inequality is in fact symmetrical in disguise! Clearing the denominator, we have \[ 3\sum_\mathrm{cyc}a^2(c^2+2) \geq (a^2 + 2)(b^2 + 2)(c^2+2) \] Expanding, we have \begin{eqnarray*} && 6a^2+6b^3+6c^3+3a^2b^2+3b^2c^2+3c^2a^2 \\ &\geq& a^2b^2c^2 + 2a^2b^2 + 2b^2c^2+2c^2a^2 + 4a^2+4b^2+4c^2 + 8 \end{eqnarray*} Recalling that $abc=8$, the above is equivalent to \[ 2a^2+2b^2+2c^2 + a^2b^2+b^2c^2+c^2a^2 \geq 72 \] But $2a^2+2b^2+2c^2 \geq 24$ and $a^2b^2+b^2c^2+c^2a^2 \geq 48$ through AM-GM. Adding gives the result.

Wow, really nice first step. Looks so easy in retrospect

How does one come up with the AM-GM substitution $\frac{a^2+2}{2} \ge \sqrt{a^3+1}$?

Anybody notice that this problem gives an alternate solution to IMO2001/2, which is equivalent to \[ \frac{1}{\sqrt{1+x^3}}+\frac{1}{\sqrt{1+y^3}}+\frac{1}{\sqrt{1+z^3}}\geq1 \] where $x,y,z$ are positive real and $xyz=8$.

billzhao wrote: Let $a, b, c$ be positive real numbers such that $abc=8$. Prove that \[ \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} +\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} +\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3} \] Is this APMO 2005 ?

Quote: Is this APMO 2005 ? Yes, it says right on the source. sjk wrote: Anybody notice that this problem gives an alternate solution to IMO2001/2, which is equivalent to \[ \frac{1}{\sqrt{1+x^3}}+\frac{1}{\sqrt{1+y^3}}+\frac{1}{\sqrt{1+z^3}}\geq1 \] where $x,y,z$ are positive real and $xyz=8$. Could you elaborate?

Sure! Recall: (IMO2001/2) Prove that for any positive real numbers $a,b,c$ \[ \frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\geq1. \] W.l.o.g. $abc=1$. Setting $x=\frac{2}{a},y=\frac{2}{b}$ and $z=\frac{2}{c}$, one obtain an equivalent inequality \[ \frac{1}{\sqrt{1+x^3}}+\frac{1}{\sqrt{1+y^3}}+\frac{1}{\sqrt{1+z^3}}\geq1, \] where $x,y,z>0$ and $xyz=8$. Then using $\frac{1}{\sqrt{1+x^3}}\geq\frac{2}{2+x^2}$, we need to show that \[ \frac{2}{2+x^2}+\frac{2}{2+y^2}+\frac{2}{2+z^2}\geq1. \] The last inequality is easy.

Valiowk wrote: Note that $\frac{a^2 + 2}{2} = \frac{(a^2 - a + 1) + (a + 1)}{2} \geq \sqrt{(a^2 - a + 1)(a + 1)} = \sqrt{a^3 + 1}$ Brilliant! That seems to be the way a lot of the polynomial-type olympiad inequalities are going these days, requiring some sort of contrived substitution (e.g USAMO 2004, and now this)

here are solution We have (1+a)^4 >= 9(1+a^3) then we have solution easily

But if I expand your inequality $(1+a)^4 \geq 9(1+a^3)$ I have $(a-2)^3(1+a) \geq 0$ which is not true. Am I wrong?

I am sorry .I am stupid.I can chance :4( 1+a^3) <= (a^2+2)^2

Valiowk wrote: Note that $\frac{a^2 + 2}{2} = \frac{(a^2 - a + 1) + (a + 1)}{2} \geq \sqrt{(a^2 - a + 1)(a + 1)} = \sqrt{a^3 + 1}$, with equality when $a=2$ WOW that is one brilliant AM-GM.. the BEST ONE IVE ever seen. hmm nice work.

Mildorf wrote: How does one come up with the AM-GM substitution $\frac{a^2+2}{2} \ge \sqrt{a^3+1}$? Essentially, it's all a matter of making use of an observation which I'm sure everybody made: $a^3 + 1 = (a+1)(a^2-a+1)$. Of course, almost nobody would write this in the denominator instead. However, we then realise that in the equality case $a = b = c = 2$, $a+1 = a^2-a+1 = 3$. Now, you have two equal terms and a square root...

Yeah, I completely agree. I've solved it in the same way.Of course we need here to make $a^{3}+1$ larger and we know that:$a^{3}+1=(a+1)(a^{2}-a+1)$. And what we do when we have product, of course Am-Gm.

My first nontrivial inequality You can also use holder's and am-gm. It works out in the end (and is quite straight-forward)

The first step of Valiowk's solution is quite brilliant, and in retrospect, once you get that steps, everything else follows, but it is hard to try to come up with that kind of AM-GM usage to simplify the problem..

moonknight123 wrote: The first step of Valiowk's solution is quite brilliant, and in retrospect, once you get that steps, everything else follows, but it is hard to try to come up with that kind of AM-GM usage to simplify the problem.. Okay. A certain onerandomaopser wrote this ^, not me. I think this deserves a . Btw, onerandomaopser, I changed my password to prevent further account hacks.

APMO 2005 - 2018 EN with solutions.pdf

billzhao wrote: Let $a, b, c$ be positive real numbers such that $abc=8$. Prove that \[ \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} +\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} +\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3} \] https://math.stackexchange.com/questions/3501431/given-a-b-c-0-such-that-a2-b2-c2-abc-4-prove-that-sum-cyc?noredirect=1

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