orl wrote: The non-negative numbers $ x_1, x_2, \ldots, x_5$ satisfy $ \sum_{i = 1}^5 \frac {1}{1 + x_i} = 1$. Prove that $ \sum_{i = 1}^5 \frac {x_i}{4 + x_i^2} \leq 1$. $ \sum_{i = 1}^5 \frac {x_i}{4 + x_i^2} \leq 1\Leftrightarrow\sum_{i = 1}^5 \left(\frac{1}{5}-\frac {x_i}{4 + x_i^2}+\frac{3}{4}\left(\frac{1}{1+x_{i}}-\frac{1}{5}\right)\right)\geq0\Leftrightarrow$ $ \Leftrightarrow\sum_{i=1}^5\frac{(x_i-4)^2(x_i+4)}{(4+x_i^2)(1+x_i)}\geq0.$

Nice problem: Set: $ a_i = \frac {1}{1 + x_i}$ then $ x_i = \frac {1 - a_i}{a_i} = \frac {\sum_{j \neq i}{a_j}}{a_i}$ Now the i-th fraction of the LHS becomes: $ \frac {{a_i}\sum_{j \neq i}{a_j}}{4{a_i}^2 + (\sum_{j \neq i}{a_j})^2} = \frac {{a_i}\sum_{j \neq i}{a_j}}{\sum_{j \neq i}{({a_i}^2 + {a_j}^2)} + 2\sum_{k,j \neq i, j < k}{a_ka_j}}$ Now because $ {a_i}^2 + {a_j}^2 \geq 2a_ia_j$, the i-th fraction is less or equal to: $ \frac {{a_i}\sum_{j \neq i}{a_j}}{\sum_{j \neq k}{a_ka_j}}$ Adding the above for $ i = 1,2,3,4,5$, we get the desired!

Let $y_i = \frac 1{1+x_i}$ for simplicity. Then I claim $$\sum_{i=1}^5 \frac{x_i}{4+x_i^2} = \sum_{i=1}^5 \frac{y_i - y_i^2}{5y_i^2 - 2y_i+1} \geq \sum_{i=1}^5 \frac 34 y_i + \frac 1{20} = \frac 34 + \frac 1{20} \cdot 5 = 1.$$Indeed, the second inequality simplifies to $$(1-5y_i)^2(3y_i+1) \geq 0,$$which is obvious as the $y_i$ are positive, so we're done.

Is Chebyshev inequality possible here?