this has surely been posted on the forum before, because I have seen it it's trivial indeed using cauchy.

Lets denote with E that expression! It's easy to see that we have: $ E \geq \frac{ (\sum a_{1})^{2}}{\sum a_{1}+ \sum b_{1}} =\frac{1}{2}$ So the minimum is $\frac{1}{2}$. right?

We can easily generalize this: Let n be a positive integer and $a_{1},a_{2}, \ldots,a_{n},b_{1},b_{2},\ldots, \b_{n}$ be 2n positive real numbers so that $ \sum a_{1}= \sum b_{1}=k$ with k>0 and q $\in N$ ($ q\geq2$). Find the minimal value of: $ \sum \frac{a_{1}^{q}}{(a_{1}+b_{1})^{q-1}}$. In the same way we get that: $ \sum \frac{a_{1}^{q}}{(a_{1}+b_{1})^{q-1}} \geq \frac{( \sum a_{1})^{q}}{( \sum a_{1}+ \sum b_{1})^{q-1}}= \frac{k^{q}}{(2k)^{q-1}}$. for k=1 and q=2 we have this problem. cheers!

This was problem 4...But a TST problem with an one-line solution, as noticed by Valentin, it is maybe really too easy... Pierre.

a1 a1b1 a1+b1 ______=a1- ______>=a1- ________ a1+b1 a1+b1 4 when u sum u obtain 1/2 so easy man sorry guys

pbornsztein wrote: Let $n$ be a positive integer, and $a_1,...,a_n, b_1,..., b_n$ be $2n$ positive real numbers such that $a_1 + ... + a_n = b_1 + ... + b_n = 1$. Find the minimal value of $ \frac {a_1^2} {a_1 + b_1} + \frac {a_2^2} {a_2 + b_2} + ...+ \frac {a_n^2} {a_n + b_n}$. It's really easy. But let's try another solution. Let $ \displaystyle X=\frac {a_1^2} {a_1 + b_1} + \frac {a_2^2} {a_2 + b_2} + \cdots + \frac {a_n^2} {a_n + b_n}$ and define it's dual $ \displaystyle Y=\frac {b_1^2} {a_1 + b_1} + \frac {b_2^2} {a_2 + b_2} + \cdots + \frac {b_n^2} {a_n + b_n}$. Then $\displaystyle X-Y = \sum_{i=1}^n \frac {a_i^2-b_i^2} {a_i + b_i} = \sum_{i=1}^n (a_i-b_i) = 0$. On the other hand $\displaystyle X+Y = \sum_{i=1}^n \frac {a_i^2+b_i^2} {a_i + b_i} \geq \sum_{i=1}^n\frac{1}{2}(a_i+b_i) = 1$. Therefore $X=Y \geq \frac{1}{2}$.

Yes, it has been solved like that by some of the contestants. Most of the solvers did with Cauchy. One uses partial derivatives for a 3 pages solution Other uses convexity for a function with more than one variable to get a wrong solution. And many didn't solve it at all... Pierre.

pbornsztein wrote: Yes, it has been solved like that by some of the contestants. Most of the solvers did with Cauchy. One uses partial derivatives for a 3 pages solution Other uses convexity for a function with more than one variable to get a wrong solution. And many didn't solve it at all... Pierre. LOL By the way, how many are allowed to take TST? In US, only USAMO winners (max. 12) can take TST. Look at problem 11 of http://www.bath.ac.uk/~masgcs/ukimo2004/team_sheet2.pdf and at the top of the solutions http://www.bath.ac.uk/~masgcs/ukimo2004/ts2.pdf The same problem was being used for preparation last year by the top nation at IMO 2003 And it was being reused by UK and France for IMO 2004 preparation and TST!

Another solution: It's easy to verify that $\displaystyle \frac{x^2}{x+y} \geq \frac{3}{4}x - \frac{1}{4}y$ for $x,y>0$. So $\displaystyle \sum_{i=1}^n \frac {a_i^2} {a_i + b_i} \geq \frac{3}{4} \sum_{i=1}^n a_i - \frac{1}{4} \sum_{i=1}^n b_i = \frac{1}{2}$. Pierre, did anybody use this method?

There were around 25 contestants (I do not remember exactly) : The main part comes from our correspondance program. We add some from the 'Concours gnral' (which is now a big problem with a lot of questions for a lot of pages, and absolutely not an olympiad type competition), and some students from the lyce Louis-le-Grand, which is one of the best lycees in France (and because the test was organised there). From what I've said, there is no surprise that each of the 6 members of the french Imo-team were coming from our correspondance program (the others had never seen an olympiad problem). Pierre.

math_sipo wrote: a1 a1b1 a1+b1 ______=a1- ______>=a1- ________ a1+b1 a1+b1 4 when u sum u obtain 1/2 so easy man sorry guys you can use latex to write up your stuff. Check out the latex tutorial in the latex help.

1/2 .easy

By Titu's lemma we have that: $$\sum_{k=1}^{n} \dfrac{a_k^2}{a_k+b_k} \ge \dfrac{\left(\sum_{k=1}^{n} a_k \right)^2}{\sum_{k=1}^{n} a_k+\sum_{k=1}^{n} b_k}=\dfrac{1}{2}$$

This problem must be a joke. Using Cauchy-Schwarz we are done! I wonder why this problem was problem 4.....