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.
There are \(16\) cities in the kingdom. Prove that it is 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 five roads coming out of each city.
Today’s topic is inequalities, expressions like \(a\geq b\), or \(a>b\). There are certain rules for operating inequalities: one can subtract the same number from both sides of the inequality, namely if \(a\geq b\), then \(a-b \geq 0\). If \(a \geq b\) and \(b\geq c\), then \(a\geq c\). If a number \(c\geq 0\), then from \(a\geq b\) it follows that \(ac \geq bc\). However, in case of multiplication by a negative number \(c\leq 0\), the inequality sign reverses: from \(a\geq b\) it follows that \(ac \leq bc\). One should also remember that the square of any real number is non-negative.