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If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$

I know that $$a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\ge ab+bc+ac$$

So we could prove that $$a+b+c+3\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$

I.e. $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3$$

By using AM-GM we see that we could prove that: $$2(a+b+c)\le a+b+c+3\Rightarrow a+b+c\le3$$

So this didn't work.

user26486
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    Why would you ever want to prove something this insane. What is this a math competition problem? – Ethan Splaver Dec 27 '13 at 22:46
  • Yes, it's a problem from the European mathematical cup of 2013. – user26486 Dec 27 '13 at 22:48
  • I don't see how you use AM-GM in your last step. Could you clarify on that? – Ragnar Dec 27 '13 at 22:50
  • $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le (a+b)+(b+c)+(c+a)=2(a+b+c)$$ – user26486 Dec 27 '13 at 22:53
  • Probably, one of your inequalities isn't sharp enough, and because of that you cannot ever prove the question. – Ragnar Dec 27 '13 at 22:53
  • Do you mean that the inequalities that I've derived aren't sharp enough? And that we could never prove the derived ones? I don't understand. – user26486 Dec 27 '13 at 22:55
  • Suppose we want to show $5\geq 4$. I could say: 'Well, we know that $5\geq 3$, so now we only have to show that $3\geq 4$.' but I will never succeed in doing that. – Ragnar Dec 27 '13 at 22:57
  • If we want to show that $a\ge c$ and we know that $a\ge b$, we will succeed in proving that $a\ge c$ if we find out that $b\ge c$. I've said in my solution that we could prove that $b\ge c$, but we don't know if that's true. If we proved that, we'd know that $a\ge c$. That's the way inequalities are usually proven. – user26486 Dec 27 '13 at 23:01
  • In your case, if we proved that $3\ge 4$, then we'd know that $5\ge 4$ if $5\ge 3$. But in this case this isn't true. – user26486 Dec 27 '13 at 23:04
  • In general, you cannot solve inhomogeneous inequalities without using some given conditions. – Ragnar Dec 27 '13 at 23:30
  • How do you know that? How to prove it? And an inequality is given as a condition. I think we could get quite a simple condition somehow by rearranging the inequality. By "condition" I think you mean an equality like $a+b+c=3$ or $abc=1$. And I don't know how you've deduced that one can't prove an inhomogeneous inequality without a condition. – user26486 Dec 27 '13 at 23:37
  • @Ragnar Not sure what you mean. I don't see why we cannot have non-homogeneous inequalities without additional conditions, for e.g. it is easy to see for all reals $x^2+y^2 + 2 \ge 2(x+y)$ holds. I am sure one can come up with any number of other examples. – Macavity Dec 28 '13 at 07:40
  • What do you mean by "$LHS$" and "$RHS$"? We can show that $a+b+c+3=M\ge N=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$, because $M$ is either reduced or equal to $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2=L$. If a reduced-or-equal expression is greater than or equal to $N$, then the expression that was reduced has to be greater than or equal to $N$ too. $N\le O$, so if we prove that $O\le M$, then we'll know that $N\le M$ and thus we'll know that $M\ge N$. We know that $L\ge M$, so $L\ge N$. So we can just prove that $O\le M$ and everything will be proven. – user26486 Dec 28 '13 at 15:09
  • And you've shown with a counterexample that we can't prove that $a+b+c\le 3$ because it's not true, so we can try proving the other inequality I've written in my steps: $a+b+c+3\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. – user26486 Dec 28 '13 at 15:11
  • Obviously I know that "$LHS$" stands for "left hand side" and "$RHS$" stands for the right. But $LHS$ and $RHS$ of which result? Tell me the expressions that $LHS$ and $RHS$ represent in your case. – user26486 Dec 28 '13 at 18:15
  • http://i.stack.imgur.com/wkHCC.png

    We know that M is behind L and that N is behind O (on the x-axis, according to their values). But we don't know the connection between the segments ON and ML. Now, notice that if O is behind M, then N is behind M and so N is behind L too. So we only have to prove that O is behind M. Just like I said before. Notice that everytime I said "behind" I meant "behind or at the same point", because equalities can hold too.

     – user26486 Dec 28 '13 at 18:48
  • I didn't understand what you meant by $LHS$ and $RHS$ because you stated false statements. $LHS\ge M$, not $M\ge LHS$. And that's where the flaw of your arguments is. Think about it again now. – user26486 Dec 28 '13 at 18:55
  • I've already pointed out that one case of this route doesn't work, i.e. it's not true that $a+b+c\le 3$, but it doesn't prove that we can't solve the inequality using the route. We can still try proving that $a+b+c+3\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. – user26486 Dec 28 '13 at 20:53
  • @Macavity So that's a counterexample and now we know that we can't prove that $a+b+c+3\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$ because it isn't true. I forgot to try out the $(4,1,1)$ triplet for this inequality, I only used it for proving that $a+b+c \le 3$ isn't true. – user26486 Dec 29 '13 at 02:12

2 Answers2

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The given condition is equivalent to: $$\sum_{cyc}\frac{c-ab}{1+a+b} \ge 0$$ $$\implies \sum_{cyc}\frac{c(1+a+b)-ab}{1+a+b} \ge \sum_{cyc}\frac{c(a+b)}{1+a+b} \qquad \text{adding to both sides}$$ $$\implies \sum_{cyc}c \ge \sum_{cyc}\frac{c(a+b)+ab}{1+a+b} = (ab+bc+ca)\sum_{cyc}\frac{1}{1+a+b} \tag{1}$$

By Cauchy-Schwarz, we also have $$\left(\sum_{cyc}\frac{1}{1+a+b}\right)\left(\sum_{cyc}(c+ca+bc)\right)\ge \left(\sum_{cyc} \sqrt{c} \right)^2 \tag{2} $$

Using this in $(1)$, we have: $$a+b+c \ge (ab+bc+ca)\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{a+b+c+2(ab+bc+ca)}$$

Cross multiplying and expanding we get, $$(a+b+c)^2 + 2(ab+bc+ca)(a+b+c)\ge (ab+bc+ca)(a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca})$$

Expanding the first term, cancelling part of the second term with RHS and dividing throughout by $ab+bc+ca$, we get the result.

Macavity
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Here are some thought I have, but no proof (yet):
Both the condition and the inequality to be proven have equality when $(a,b,c)=(x,x,x)$ and when $(a,b,c)=(1,1,4)$ (or permutations). We know that the inequalities of means only have equality when the term we apply them to are equal. Thus, we can never use AM-GM on $a$, $b$ and $c$, because $(1,1,4)$ wouldn't give equality and thus the inequality is to lossy.
If we apply AM-GM on $f(a,b,c)$, $g(a,b,c)$ and $h(a,b,c)$, we want $f(1,1,4)=g(1,1,4)=h(1,1,4)$, $f(1,4,1)=g(1,4,1)=h(1,4,1)$ and the same for $(1,1,4)$. I think it is not easy to find functions that satisfy this (apart from $f=g=h$, but then, you won't get any information from the inequality).
We can thus only use the given inequality and use rewriting of terms. Maybe we can also use Cauchy-Schwarz, because it has not-so-trivial cases for equality.

Ragnar
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