Problems

Age
Difficulty
Found: 1866

Let \(k\) be a natural number, prove the following inequality. \[\frac1{k^2} > \frac1{k} - \frac1{k+1}.\]

Show that if \(a\) is a positive number, then \(a^3+2 \ge 2a \sqrt{a}\).

The numbers \(a\), \(b\) and \(c\) are positive. By completing the square, show that \[\frac{a^2}4 + b^2 + c^2 \ge ab-ac+2bc.\]

Let \(m\) and \(n\) be natural numbers such that \(m>n\). Show that: \[\frac1{n^2} + \frac1{(n+1)^2} + \frac1{(n+2)^2} + \dots + \frac1{m^2} > \frac1{n} - \frac1{m}.\]

The numbers \(a,b,c\) are positive. Show that: \[\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b} \ge a +b+c.\]

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 \(t(n)\)?

Consider the following sum: \[\frac1{1 \times 2} + \frac1{2 \times 3} + \frac1{3 \times 4} + \dots\] Show that no matter how many terms it has, the sum will never be larger than \(1\).

Cut a \(7\times 7\) square into \(9\) rectangles, out of which you can construct any rectangle whose sidelengths are less than \(7\). Show how to construct the rectangles.

There are \(16\) cities in the kingdom. Prove that it is not possible to build a system of roads in such a way that one can get from any city to any other without passing through more than one city on the way, and with at most four roads coming out of each city.