oVlad wrote: Let $a_1,\ldots,a_{n+1}$ be positive real numbers satisfying $1/(a_1+1)+\cdots+1/(a_{n+1}+1)=n$. Prove that \[\sum_{i=1}^{n+1}\prod_{j\neq i}\sqrt[n]{a_j}\leqslant\frac{n+1}{n}.\] The following inequality is much more stronger and also easy. Let $a_i$ be non-negative numbers such that $$\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_{n+1}+1}=n.$$Prove that:

$$\sum_{1\leq i<j\leq n+1}\sqrt{a_ia_j}\leq\frac{n+1}{2},$$but