Prove that amongst any 7 different numbers it is always possible to choose two of them, \(x\) and \(y\), so that the following inequality was true: \[0 < \frac{x-y}{1+xy} < \frac{1}{\sqrt3}.\]
The number \(n\) is natural. Show that: \[\frac1{\sqrt{1}} +\frac1{\sqrt{2}}+ \frac1{\sqrt{3}} + \dots +\frac1{\sqrt{n}} < 3 \sqrt{n+1} -3.\]
If \(n\) is a positive integer, we denote by \(s(n)\) the sum of the divisors of \(n\). For example, the divisors of \(n=6\) are \(1,2,3,6\), so \(s(6)=1+2+3+6=12\). Prove that, for all \(n\geq1\), \[s(1)+s(2)+\cdots+s(n)\leq n^2.\] Denote by \(t(n)\) is instead the sum of the squares of the divisors of \(n\) (e.g., \(t(6)=1^2+2^2+3^2+6^2=50\)), can you find a similar inequality for \(t(n)\)?