danepale wrote: Let $n \ge 2$ be an integer, and let $x_1, x_2, \ldots, x_n$ be reals for which: (a) $x_j > -1$ for $j = 1, 2, \ldots, n$ and (b) $x_1 + x_2 + \ldots + x_n = n.$ Prove that $$ \sum_{j = 1}^{n} \frac{1}{1 + x_j} \ge \sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2} $$and determine when does the equality occur. $ \sum_{j = 1}^{n} \frac{1}{1 + x_j}-\sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2}=\sum_{j=1}^n\frac{1-x_j}{(1+x_j)(1+x_j^2)}=$ $=\sum_{j=1}^n\left(\frac{1-x_j}{(1+x_j)(1+x_j^2)}+\frac{1}{4}(x_j-1)\right)=\sum_{j=1}^n\frac{(x_j-1)^2(x_j^2+2x_j+3)}{4(1+x_j)(1+x_j^2)}\geq0$ The equality occurs for all $x_j=1$.

This is waaay too easy for MEMO. $$LHS = \sum_{cyc} \frac{1}{x_1+1} \underset{CSB}{\geq} \frac{n^2}{n+x_1+...+x_n} = \frac{n^2}{2n} = \frac{n}{2}$$$$RHS = \sum_{cyc} \frac{x_1}{x_1^2+1} \leq \sum_{cyc} \frac{1}{2} = \frac{n}{2}$$$$LHS \geq \frac{n}{2} \geq RHS$$The first inequality works because the denominators are all positive and the second one similarly from $(x-1)^2 \geq 0$. Equality is achieved when $x_i = 1, \forall i \leq n$

You could also do tangent inequality. For $f(x)= {1-x\over (1+x)(1+x^2)}$ we have tangent $y=-{1\over 4}x+{1\over 4}$ at $T(1,0)$. It is easy to see $f(x)\geq y$ for all $x>-1$ and thus conclusion.

danepale wrote: Let $n \ge 2$ be an integer, and let $x_1, x_2, \ldots, x_n$ be reals for which: (a) $x_j > -1$ for $j = 1, 2, \ldots, n$ and (b) $x_1 + x_2 + \ldots + x_n = n.$ Prove that $$ \sum_{j = 1}^{n} \frac{1}{1 + x_j} \ge \sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2} $$and determine when does the equality occur. $$ \sum_{j = 1}^{n} \frac{1}{1+x_j} \ge\sum_{j = 1}^{n} \frac{n^2}{n + \sum_{j = 1}^{n} x_j} =\frac{n}{2}=\sum_{j = 1}^{n} \frac{|x_j|}{2|x_j|}\ge \sum_{j = 1}^{n} \frac{|x_j|}{1+x^2_j}\ge\sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2} $$

Tangent line method, with $\frac{1}{1+x}-\frac{x}{1+x^2}+\frac{1}{4}(x-1) \ge 0$ for all $x >-1$.

WOW, I suck at inequalities.. Let me present a very long solution.. We'll proceed by induction on $n$. Base step: The case $n=2$ is easy to prove by multiplying out and plugging in $x_1=2-x_2$ and taking derivatives to find its global minimum. The equality case is $x_1=x_2=1$. Induction Hypothesis: The inequality holds for $n-1$ and equality happens at $x_1=x_2=\dots=x_{n-1}=1$. Induction Step: Assume any of $x_1,x_2,\dots,x_{n}$ was $1$. The inequality reduces to an equivalent form for $n-1$ with which we're done by I.H. with the equality case being $a_1=a_2=\dots=a_{n}=1$ again. Now suppose $a_i \neq 1$ for all $1 \leq i \leq n$. Then \begin{align*} \sum_{j=1}^{n} \frac{1}{1+x_j} &\geq \sum_{j=1}^{n} \frac{x_j}{1+x_j^2} \\ \iff \sum_{j=1}^n \left( \frac{x_j}{x_j^2-1}-\frac{1}{x_j^2-1} \right) &\geq \sum_{j=1}^{n} \frac{x_j}{1+x_j^2} \\ \iff \sum_{j=1}^n x_j \cdot \left(\frac{1}{x_j^2-1}-\frac{1}{x_j^2+1} \right) &\geq \sum_{j=1}^n \frac{1}{x_j^2-1} \\ \iff \sum_{j=1}^n \frac{2x_j}{x_j^4-1} &\geq \sum_{j=1}^n \frac{1}{x_j^2-1} \\ \iff \sum_{j=1}^n \frac{- \left(x_j^2+1 \right)^2}{x_j^4-1} &\geq 0 \\ \iff \sum_{j=1}^n \frac{x_j^2+1}{x_j^2-1} &\leq 0 \\ \iff \sum_{j=1}^n \frac{1}{1-x_j^2} &\geq \frac{n}{2} \\ \iff \sum_{j=1}^n \left(\frac{1}{1-x_j}+\frac{1}{1+x_j} \right) &\geq n \end{align*}Now by Cauchy-Schwarz we get \[ \sum_{j=1}^n \frac{1}{1+x_j} \geq \frac{n^2}{n+n}=\frac{n}{2} \]and \[ \sum_{j=1}^n \frac{1}{1-x_j} \geq \frac{(n-1)^2}{n-1-(n-x_n)} + \frac{1}{1-x_n} = \frac{n^2-2n}{x_n-1} > \frac{n^2-2n}{n-1}. \]Now \[ \frac{n^2-2n}{n-1} \geq \frac{n}{2} \iff 2n-4 \geq n-1 \iff n \geq 3. \]Thus, we're done. And no equality case, if none of the numbers are $1$, therefore the equality case remains at all numbers to equal $1$. Also note that all of those equations above are indeed equivalent by the conditions we have. $\hfill \square$

Axiomatical wrote: This is waaay too easy for MEMO. Can you justify this assessment? 20 of the 60 participants obtained zero marks on this problem including three bronze medal winners who had to solve two other problems to get their medals. The weakest countries gained one medal, and only one country gained six medals. This all seems about right to me.

I must admit, I was surprised myself when I saw the results. It was the worst solved problem (besides, of course, geometry). I come from Croatia, so I might be biased as our country did do really well on MEMO this year, but this is a routine problem compared to our national olympiad problems. I would also like to note that besides this year and the past year, MEMO has always had hard inequality problems and this is a cakewalk in comparison.

$$ \sum_{j = 1}^{n} \frac{1}{1 + x_j} \ge \sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2} $$$$\frac{1}{2} \ge \frac{x_j}{1 + x_j^2}$$$$\frac {n}{2} \ge \sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2} $$$$ (\sum_{j = 1}^{n} \frac{1}{1 + x_j})( \sum_{j = 1}^{n}(1+x_j)) \ge n^2 $$$$ (\sum_{j = 1}^{n} \frac{1}{1 + x_j})(2n) \ge n^2 $$$$ (\sum_{j = 1}^{n} \frac{1}{1 + x_j}) \ge \frac{n}{2}$$$$ (\sum_{j = 1}^{n} \frac{1}{1 + x_j}) \ge \frac {n}{2} \ge \sum_{j = 1}^{n} \frac{x_j}{1 + x_j^2} $$