this one is very hard without the following method (e.g. expanding it out ) (BTW this isnt my proof) x <= 1+x^2 / 2, etc. so it suffices to prove, after the substitution a=1+x^2, b=1+y^2, c=1+z^2 that a/c+(b/2) + b/a+(c/2) + c/b+(a/2) >= 2, which is easy

You said "... which is easy..."Why?After applying the AM/GM twice we have 3 {abc/[(2c+b)(2a+c)(2b+c)]}^1/3 which we should prove is >eq 2.How does one prove that?

I'll try to clarify. As $x \leq \frac{1 + x^2}{2}$ we have \[\sum \frac{1 + x^2}{1 + y + z^2} \geq \sum \frac{2(1 + x^2)}{(1 + y^2) + 2(1 + z^2)}.\] Denoting $1 + x^2 = a$ and so on we have to prove that \[\sum \frac{a}{b + 2c} \geq 1\] but Cauchy tells us \[\sum \frac{a}{b + 2c} \sum a(2b + c) \geq \left(\sum a\right)^2\] and as \[\left(\sum a\right)^2 \geq 3(ab + bc + ca) = \sum a(2b + c)\] we have the result.

Posted before. The essential idea is to make use of the inequality $\frac{x^2+1}{2}\ge x$ and then Cauchy-Schwarz.

it is not new. It was posted in Junior Balkan Mathematical Olympiad 2003 and it was proposed by Laurentiu Panaitopol.

mecrazywong wrote: Posted before. The essential idea is to make use of the inequality $\frac{x^2+1}{2}\ge x$ and then Cauchy-Schwarz. Can you show?

Well, I'm tired of saying it again and again ,but the initial inequality reduces to this well-known one: $\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1$ which can be easily solved by using Cauchy or by AM-GM.

I still can't see how..

$\frac{1+x^2}{1+y+z^2}\geq\frac{1+x^2}{1+\frac{1+y^2}{2}+z^2}=\frac{2(1+x^2)}{1+y^2+2(1+z^2)}$ I think you can see now...

OK,thanks! Are there any other solutions?

Man, i have already told you. Another solution is to make the substitutions $2a+b=m, 2b+c=n, 2c+a=p$ and after that the inequality reduces to a direct application of AM-GM.

Cezar Lupu wrote: Well, I'm tired of saying it again and again ,but the initial inequality reduces to this well-known one: $\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1$ which can be easily solved by using Cauchy or by AM-GM. Can any one show me how to solve it using AM-GM?

After those substitutions, I can't get AM-GM to work.

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=64654 . (for $\frac a{2b+c} + \frac b{2c+a} + \frac c{2a+b} \ge 1$, and a bit more )

perfect_radio wrote: After those substitutions, I can't get AM-GM to work. a,b,c>0...so is m,n,p>0 so when you substitute those values you will have 9a= 4m-2n+p....etc. putting them together in the inequality you'll have something like this.... 4(r+1/r)+4(s+1/s)+4(t+1/t)>=.... ; where r=m/n, s=n/p, t=p/m...etc. AM-GM from there

Is the condition $ x,y,z>-1$ necessary? I can't seem to see where it is used.

Yes, it is: taking $ z = 0$ and $ y$ just slightly smaller than $ - 1$ allows the left-hand side to take arbitrarily big negative values. It appears that it's used in the proof in the form of the (implicit) assumption that the fractions in the question are positive.

Arne wrote: I'll try to clarify. As $ x \leq \frac {1 + x^2}{2}$ we have \[ \sum \frac {1 + x^2}{1 + y + z^2} \geq \sum \frac {2(1 + x^2)}{(1 + y^2) + 2(1 + z^2)}.\] Denoting $ 1 + x^2 = a$ and so on we have to prove that \[ \sum \frac {a}{b + 2c} \geq 1\] but Cauchy tells us \[ \sum \frac {a}{b + 2c} \sum a(2b + c) \geq \left(\sum a\right)^2\] and as \[ \left(\sum a\right)^2 \geq 3(ab + bc + ca) = \sum a(2b + c)\] we have the result. Where is the (implicit) assumption that the fractions are positive ? The only fact used in the above solution is $ x \leq \frac {1 + x^2}{2}$ , which is true for all real $ x$ . ($ a,b,c$ are clearly positive for any real $ x,y,z$ ). I didn't get where the condition is used.

The logic of the very first step is as follows: "$ 0 \neq a \leq b \neq 0$ and therefore $ \frac{c}{a} \geq \frac{c}{b}$." This is not true for arbitrary reals -- in particular, it can fail if $ a$ is negative. Another way of saying the same thing: just go through the whole proof with $ z = 0$ and $ y = -1 - \varepsilon$ for some very small $ \varepsilon > 0$ and see which statements fail.

Shanku wrote: Arne wrote: I'll try to clarify. As $ x \leq \frac {1 + x^2}{2}$ we have \[ \sum \frac {1 + x^2}{1 + y + z^2} \geq \sum \frac {2(1 + x^2)}{(1 + y^2) + 2(1 + z^2)}.\] Denoting $ 1 + x^2 = a$ and so on we have to prove that \[ \sum \frac {a}{b + 2c} \geq 1\] but Cauchy tells us \[ \sum \frac {a}{b + 2c} \sum a(2b + c) \geq \left(\sum a\right)^2\] and as \[ \left(\sum a\right)^2 \geq 3(ab + bc + ca) = \sum a(2b + c)\] we have the result. Where is the (implicit) assumption that the fractions are positive ? The only fact used in the above solution is $ x \leq \frac {1 + x^2}{2}$ , which is true for all real $ x$ . ($ a,b,c$ are clearly positive for any real $ x,y,z$ ). I didn't get where the condition is used. JBL wrote: The logic of the very first step is as follows: "$ 0 \neq a \leq b \neq 0$ and therefore $ \frac {c}{a} \geq \frac {c}{b}$." This is not true for arbitrary reals -- in particular, it can fail if $ a$ is negative. Another way of saying the same thing: just go through the whole proof with $ z = 0$ and $ y = - 1 - \varepsilon$ for some very small $ \varepsilon > 0$ and see which statements fail. Just now my eyes opened ! I posted stupid question. Sorry. The condition is to - [1] avoid zero in the denominators. [2] keep the fractions in the positive zone only. Cheers

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can also be proved above like this $\sum_{\text{cyc}}\frac{a^2}{(2c+b)a} $ $\ge \frac{(a+b+c)^2}{3(ab+bc+ac)}\ge 1$

Let $a$, $b$ and $c$ be positive numbers. Prove that $$\frac{a^2+1}{ab+1}+\frac{b^2+1}{bc+1}+\frac{c^2+1}{ca+1}\geq \frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1} ,$$$$\frac{a^2+1}{(ab+1)^2}+\frac{b^2+1}{(bc+1)^2}+\frac{c^2+1}{(ca+1)^2}\geq \frac{9}{2(abc+2)}.$$Jichen p/6483274195 , p/6481977917

Iris Aliaj wrote: Let $x, y, z > -1$. Prove that \[ \frac{1+x^2}{1+y+z^2} + \frac{1+y^2}{1+z+x^2} + \frac{1+z^2}{1+x+y^2} \geq 2. \]Laurentiu Panaitopol $$\sum_{cyc} \frac{1+x^2}{1+y+z^2} \geq \frac{(3+\sum_{cyc} x^2)^2}{\sum_{cyc} ((1+x^2)(1+y+z^2)} $$$$\frac{(3+\sum_{cyc} x^2)^2}{\sum_{cyc} ((1+x^2)(1+y+z^2)}\geq 2$$$$\iff$$$$\sum_{cyc} (x-1)^2+\sum_{cyc} (x^2-y)^2\geq 0$$

Here is my solution i used the main idea and simply solved so yeah i did it and i know above are also correct solution i am sending here to make this threat a life(Active) a bit Main Idea is here to see that with the help of AM-GM that \[(y^2+1)\geq 2y\]\[(z^2+1)\geq 2y\]\[(1+x^2)\geq 2x\]we can use this in main problem \[\sum_{cyc}\frac{1+x^2}{1+y+z^2}\geq \sum_{cyc}\frac{1+x^2}{1+\frac{1+y^2}{2}+z^2}=\sum_{cyc}\frac{2(1+x^2)}{(1+y^2)+2(1+z^2)}\]So we can see that by again using that \[\implies \sum_{cyc}\frac{2(1+x^2)}{(1+y^2)+2(1+z^2)}\geq \frac{4x}{2y+4z}=\frac{2x}{y+2z}\]So its suffice to show \[\implies\sum_{cyc}\frac{x}{y+2x}\geq 1\]Now we can use cauchy inequlity it makes thing simple that i suppose so \[\implies\left(\sum_{cyc}\frac{x}{y+2z}\right)\left(\sum_{cyc}x(y+2z)\right)\geq (x+y+z)^2\]So again it suffice to show that \[\implies(x+y+z)^2\geq \sum_{cyc}x(y+2z)=3(xy+yz+zx)\]\[\implies x^2+y^2+z^2\geq xy+yz+zx\]So its simple AM-GM and we are done Demon Killed mission accomplished.

\[\implies \sum_{cyc}\frac{2(1+x^2)}{(1+y^2)+2(1+z^2)}\geq \frac{4x}{2y+4z}=\frac{2x}{y+2z}\]This step is wrong @above

Other way to prove the below inequality $$\sum_{cyc} \frac{a}{b+2c} \ge 1$$Is to notice that the expression is equivalent to $$ S = \sum_{cyc} \frac{a^2}{ab+2ac}$$By C.S Engel or Titu's lemma, we have $$ S \ge \frac{(a+b+c)^2}{3(ab+bc+ca)} \ge 1$$Which is easy to prove.