OKay.. cause today is my birthday and I promissed myself that at least on my birthday I will stop solving math problems.. I will just ask you if you used the oldie but goldie trick a=x/y, b=z/x and c=y/z because their product will be 1. hope that this is how is made! (I am not solving anything today )

1st of all happy birthday Lagrangia! sencond of all - this inequality was given at a Romanian Test for 8-th graders (last school class before entering high-school in Romania), and obviously no-one did it. I've tried it then, using all elementary (and nice) ideas I could possible think of. Didn't work. Last year I found out that it was also used in a Bulgarian TST, and I succeded in solving it using the "brute force" (which they actually have in their original solution). Besides that, like that problem from Iran 1996, it is very unlikey that one can find a nice solution ...

For the trick I prefer a=x^2/yz, b=y^2/zx, c=z^2/xy instead since it doesn't cause the loss of symmetry..

Well,we can use a well-known method here.Take $f(a,b,c)=RHS-LHS$. Assume that $c=max \lbrace{a,b,c}\rbrace$ then we have: f(a,b,c)-f( \frac{a+b}{2}, \frac{a+b}{2},c)= (a-b)^2(\frac{1}{(2+a)(2+b)(4+a+b)}-\frac{1}{(1+a+c)(1+b+c)(2+2c+a+b)}) \geq 0 since c \geq 1 Now,prove it for f(a,a,b).I have expanded it and this is really disgusting . However,the last result is (a-1)^2(2a^4+9a^3+9a^2+4a+3) \geq 0.

I just wanted to mention the appearance of the inequality with solution in two sources: 1. Titu Andreescu, Vasile Cîrtoaje, Gabriel Dospinescu, Mircea Lascu, Old and New Inequalities, Zalau: GIL 2004, problem 99. 2. American Mathematics Competitions: Mathematical Olympiads 1997-1998: Olympiad Problems from Around the World, Bulgaria 21, p. 23. Both solutions are almost the same: Brute force. The inequality doesn't seem to have a better proof. The second source can be freely downloaded from http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/s-index.html EDIT: Since this website does not work properly, here are direct links to the files at the site: http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/geom-080399.ps http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/geom-080399.tex http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ineqs-080299.ps http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ineqs-080299.tex http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.dvi http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.ps http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.dvi http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.ps Darij

FOR several months I am trying to prove the general case of this damn inequality, but I did not succeded. Did you?

It seem be post some times in mathlinks. And when you Multi two hand side,it's very clearly,not difficult. I had solve it before by this way. Other way,It can be prove immediate by the generel: $\sum\frac{1}{n-1+x_i}\le 1$ if $\prod{x_i}=1$ And then: $\sum\frac{1}{n-1+x_i} \le 1 \le \sum\frac{1}{1+(n-1)x_i}$

But $1 \ge \sum{\frac{1}{n-1+x_i}}$.

hungkhtn wrote: It seem be post some times in mathlinks. $\sum\frac{1}{n-1+x_i}\le 1$ if $\prod{x_i}=1$ And then: $\sum\frac{1}{n-1+x_i} \le 1 \le \sum\frac{1}{1+(n-1)x_i}$ Here is a proof for the first one. $f(x_{1},x_{2},...,x_{n})=\sum\frac{1}{n-1+x_i}$.As $\prod{x_i}=1$ we may assume $x_{1}\ge{1}, x_{2}\le{1}$. We shall prove that $f(x_{1},x_{2},...,x_{n})\le{f(1,x_{1}x_{2},...,x_{n})}$. And this is true because after a little computation we obtain $(1-x_{1})(x_{2}-1)(x_{1}x_{2}+(n-1)^{2})\ge{0}$ which is obviously true. So we have $f(x_{1},x_{2},...,x_{n})\le{f(1,x_{1}x_{2},...,x_{n})}\le ...\le{f(1,...,1)}=1$. QED

i have just solved by clearing denominators + muirhead symmetric inequality. but is there a short solution???????

i'm very sure that mixing variables works, but i'm not sure if it's short.

Here is solution :asumme a\geq b\geq c R.H.S-L.H.S=\sum \frac{b-1}{(2+a)(1+a+b)}\geq \frac{a+b+c-3}{(2+a)(1+a+b)}\geq 0

keira_khtn wrote: $\sum \frac{b-1}{(2+a)(1+a+b)}\geq \frac{a+b+c-3}{(2+a)(1+a+b)}$ This inequality is not trivial at all, because $\frac{c-1}{(2+b)(1+b+c)}$ is negative and can't simply change the denominator.

this is from bulgaria ninety something i guess the best solution is bruteforce put : x=a+b+c and y=ab+ac+bc and express each side in terms of these variables and then pure calculation till you get this very ugly inequality wich is true using AM-GM

Bojan Basic wrote: keira_khtn wrote: $\sum \frac{b-1}{(2+a)(1+a+b)}\geq \frac{a+b+c-3}{(2+a)(1+a+b)}$ This inequality is not trivial at all, because $\frac{c-1}{(2+b)(1+b+c)}$ is negative and can't simply change the denominator. Oh,i'm false..

i m not sure whether i m right, i found a very easy solution though look, use the classical technic , put $ a = \frac xy , b = \frac yz , c = \frac zx $ and assume that $ x \geq y \geq z $ then we have $ \sum \frac {1}{1+a+b} = \sum \frac {yz}{yz + zx + y^2} \leq \sum \frac {yz}{2yz + zx} $ since $ y^2 \geq yz $ then $ \sum \frac {yz}{2yz + zx} = \sum \frac {1}{2 + \frac xy} = \sum \frac {1}{2 + a} $ hence $ \sum \frac {1}{1+a+b} \leq \sum \frac {1}{2 + a} $ any mistake??

Al-Khawarizmi wrote: $\sum \frac {yz}{yz + zx + y^2} \leq \sum \frac {yz}{2yz + zx} $ since $ y^2 \geq yz $ any mistake?? Yes. We do have $y^2 \geq yz$ but the other, similar inequalities do not necessarily hold.

my solution is wrong, sorry!

thank you Arne, i did not see your post, you are so fast here people

I think the best solution is to use the fact that $\frac{1}{2a+1}+\frac{1}{2b+1}\geq\frac{2}{a+b+1}$

[EDIT] I applyed Jensen's incorrectly. It is funny how you get the desired result with the stupid sign mistake

Guys, you know this is a hard problem and the only known solution is brute-force, so if you think you have a short simple proof, why can't you bother check it twice before posting it? For instance, Silouan's inequality silouan wrote: $\frac{1}{2a+1}+\frac{1}{2b+1}\geq\frac{2}{a+b+1}$ is true, but doesn't help solving the initial problem. Yassin's Ezbakhe Yassin wrote: $1\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$ is true, but with $\geq$ instead of $\leq$, so it isn't too useful either. Sorry, but I just wanted to tell the truth... Darij

Ezbakhe Yassin wrote: Here is a more direct form. Hope it is correct. $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{3}{\frac1{3}(1+a+b+1+b+c+1+c+a)}$$=\frac9{3+2a+2b+2c}\leq 1$ $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\geq \frac9{3+2a+2b+2c}$ by Cauchy

silouan wrote: I think the best solution is to use the fact that $\frac{1}{2a+1}+\frac{1}{2b+1}\geq\frac{2}{a+b+1}$ Then We must prove $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\geq \frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$ But it was wrong

alekk wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that $ \frac {1}{1 + b + c} + \frac {1}{1 + c + a} + \frac {1}{1 + a + b}\leq\frac {1}{2 + a} + \frac {1}{2 + b} + \frac {1}{2 + c}$. Let $ x = a + b + c;y = ab + bc + ca$ Inequality becomes: $ \frac {x^2 + 4x + y + 3}{x^2 + 2x + y + xy} \leq \frac {12 + 4x + y}{9 + 4x + 2y}$ <=>$ \frac {2x + 3 - xy}{x^2 + 2x + y + xy} \leq \frac {3 - y}{9 + 4x + 2y}$ Note that $ x\geq3;y \geq 3;x^2 \geq 3y$ so we have $ \frac {5}{3}x^2y\geq 5x^2;\frac {x^2y}{3}\geq y^2;xy^2 \geq 9x;5xy \geq 15x;xy \geq 3y;x^2y \geq 27$ Done!

alekk wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$. Prove that $ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$. Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd=1$.Prove that $$ \frac{1}{1+b+c+d}+\frac{1}{1+c+d+a}+\frac{1}{1+d+a+b}+\frac{1}{1+a+b+c} \le\frac{1}{3+a+} +\frac{1}{3+b} + \frac{1}{3+c} + \frac{1}{3+d} $$

alekk wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$. Prove that $$ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$ Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.Prove that $$\frac{1}{3}\left(\frac{1}{\sqrt{2+a}} +\frac{1}{\sqrt{2+b}}+\frac{1}{\sqrt{2+c}}\right)^2\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \leq 1$$$$\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\le \frac{3}{4}$$here here

https://artofproblemsolving.com/community/c6h1655438p10498091: Let $a,b,c>0$ and $abc=1$. Show that $$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1} \geqslant \frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge 1$$Let $a,b,c>0$ and $abc=1$. Show that $$\frac{1}{b+c+1}+ \frac{1}{c+a+1}+ \frac{1}{a+b+1}\leq\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\leq 1.$$ sqing wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd=1$.Prove that $$ \frac{1}{1+b+c+d}+\frac{1}{1+c+d+a}+\frac{1}{1+d+a+b}+\frac{1}{1+a+b+c} \le\frac{1}{3+a} +\frac{1}{3+b} + \frac{1}{3+c} + \frac{1}{3+d} $$ here