If $a=b=c= \sqrt{2}$ then $ab+bc+ca = 6$ :S

RaMlaF wrote: If $a=b=c= \sqrt{2}$ then $ab+bc+ca = 6$ :S when $a=b=c=\sqrt{2}$ $\sum\frac{1}{a^2+1}=1\neq 2$

Yes, I guess we all see that Shyong, but there was some confusion since Omid Hatami had a typo in his post.

This problem is not hard ! Expand the hypothesis , we get : $ a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2 =1 $ Put $ ab =2cos C ; ac= 2cos B , bc =2cos A $ The ineq becomes $ cos A+cos B +cos C \leq \frac{3}{2} $

nttu wrote: This problem is not hard ! Expand the hypothesis , we get : $ a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2 =1 $ Put $ ab =2cos C ; ac= 2cos B , bc =2cos A $ The ineq becomes $ cos A+cos B +cos C \leq \frac{3}{2} $ I guess you mean $ab=\cos C$ instead of $2\cos C$

This problem has been in Iran exam in (1376,problem 4) I think it worth repeating it cause it has a nice solution with Caushy. Omid Why did you guys repeat this problem ? Solution .$ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}\le \frac{3}{2}$ so its enough to prove $(a+b+c)^2 \le (a^2+1+b^2+1+c^2+1)$ Let $x=a^2+1,y=b^2+1,z=c^2+1$ The problem is equal to : Let $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ prove $\sqrt{x+y+z} \ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{y-1}$ which follows from a simple but nice caushy $\sqrt{x+y+z} *\sqrt{(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})=3-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=1} \ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$ NTTU=I am not sure that you can use that subtitution with $Cos$

you are so cool lomos_lupin(Arman).

Thanks but There is nothing to be cool about but its strange that problem was repeated in the official exam , i am more interested in the solutions of the contestants ,Omid Would you please give one of the nice solutions?

this is not only problem that was repeated Arman there are many problems. as we see in Geometry .I think the reason that OMID HATAMI didn't send Graph problems was this. most of the problem (or maybe all of them) of graph problems was in Bondy's Book and ... $So\ Easy\ to\ get\ gold\ this\ year.$

Will there be any more rounds for the Iranian Mathematical Olympiad 2005? I know that there used to be three rounds with 6 problems in the past (with the third round very hard). Did that change? If so, that's a pity. I don't like this "algebra exam", "geometry exam", ... etc

Arne wrote: Will there be any more rounds for the Iranian Mathematical Olympiad 2005? yes,there is. Actually in iran there are four rounds.4th round is very hard and cute.

lomos_lupin wrote: NTTU=I am not sure that you can use that subtitution with $Cos$ Why aren't you sure ?

I am glad to hear that. Which round was this one?

This was 3rd round

nttu wrote: lomos_lupin wrote: NTTU=I am not sure that you can use that subtitution with $Cos$ Why aren't you sure ? Why,There exist Angles like $A,B,C$ such that $A+B+C=180$ and $ab=CosC,bc=CosA,ca=CosB$? I think you reached this Existance from $a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2=1$ but how?

lomos_lupin wrote: I think you reached this Existance from $a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2=1$ but how? Indeed, take $x = bc$, $y = ca$ and $z = ab$, then $x^2 + y^2 + z^2 + 2xyz = 1$ which implies that $x$, $y$ and $z$ are the cosines of the angles of a triangle.

If $a,b,c\ge 0 $ such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$ prove that \[{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le \frac{3}{2} \le \frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} }.\] here The following inequality is also true. If $a,b,c > 0 $ such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ prove that \[abc(a-1)(b-1)(c-1)\le 8.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216&start=80