Real numbers \(x,y\) are such that \(x^2 +x \le y\). Show that \(y^2 +y \ge x\).
Let \(\triangle ABC\) be a triangle and \(D\) be a point on the edge \(BC\) so that the segment \(AD\) bisects the angle \(\angle BAC\). Show that \(\frac{|AB|}{|BD|}=\frac{|AC|}{|CD|}\).
For non-negative real numbers \(a,b,c\) prove that \[a^3+b^3+c^3 \geq \frac{(a+b+c)(a^2+b^2+c^2)}{3}\geq a^2b+b^2c+c^2a.\]
Prove Nesbitt’s inequality, which states that for positive real numbers \(a,b,c\) we have \[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}.\]
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.
Given a series of statements enumerated by the natural numbers, the strong induction principle says the following. Suppose that
The 1st statement is true (the base case).
Whenever all statements up to and including the \(n\)th statement is true, the \((n+1)\)th statement is also true (induction step).
Then the statement is true for all natural numbers. Show that the strong induction principle works.