Let \(a,b,c >0\) be positive real numbers with \(abc \leq 1\). Prove that \[\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq a+b+c.\]
From IMO 1999. Let \(n\geq 2\) be an integer. Determine the least possible constant \(C\) such that the inequality \[\sum_{1\leq i<j\leq n} x_ix_j(x_i^2 + x_j^2) \leq C(\sum_{1\leq i\leq n}x_i)^4\] holds for all non-negative real numbers \(x_i\). For this constant \(C\) find out when the equality holds.
Proposed by USA for IMO 1993. For positive real numbers \(a,b,c,d\) prove that \[\frac{a}{b+2c+3d} + \frac{b}{c+2d+3a} + \frac{c}{d+2a+3b} + \frac{d}{a+2b+3c} \geq \frac{2}{3}.\]
Prove the \(AM-GM\) inequality for positive real numbers \(a_1\), \(a_2\), ..., \(a_n\): \[\frac{a_1+a_2+...+a_n}{n}\geq \sqrt[n]{a_1a_2...a_n}.\]
Due to Paul Erdős. Each of the positive integers \(a_1\), \(a_2\), ..., \(a_n\) is less than \(1951\). The least common multiple of any two of these integers is greater than \(1951\). Prove that \[\frac{1}{a_1} + ... + \frac{1}{a_n} < 1+ \frac{n}{1951}.\]
Show that if \(1+2+\dots+n = \frac{n(n+1)}{2}\), then \(1+2+\dots+(n+1) = \frac{(n+1)((n+1)+1)}{2}\).
Show that \(1+2+\dots+n = \frac{n(n+1)}{2}\) for every natural number \(n\).
Show that if \(1+2^1+2^2+\dots+2^{10} = 2^{11} - 1\), then \(1+2^1+2^2+\dots+2^{11} = 2^{12} - 1\).
Show that \(1+2^1+2^2+\dots+2^n = 2^{n+1} - 1\) for every natural number \(n\).
What is wrong with the following proof that “all rulers have the same length" using induction?
Base case: suppose that we have one ruler, then clearly it clearly has the same length as itself.
Assume that any \(n\) rulers have the same length for the induction hypothesis. If we have \(n+1\) rulers, the first \(n\) ruler have the same length by the induction hypothesis, and the last \(n\) rulers have the same length also by induction hypothesis. The last ruler has the same length as the middle \(n-1\) rulers, so it also has the same length as the first ruler. This means all \(n+1\) rulers have the same length.
By the principle of mathematical induction, all rulers have the same length.