Is it true that if \(b\) is a positive number, then \(b^3 + b^2 \ge b\)? What about \(b^3 +1 \ge b\)?
Show that if \(a\) is positive, then \(1+a \ge 2 \sqrt{a}\).
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 \(1\).