Problems

Geometry reminder

We call two polygons congruent if all their corresponding sides and angles are equal. Triangles are the easiest sort of polygons to deal with. Assume we are given two triangles \(ABC\) and \(A_1B_1C_1\) and we need to check whether they are congruent or not, some rules that help are:

• If all three corresponding sides of the triangles are equal, then the triangles are congruent.

• If, in the given triangles \(ABC\) and \(A_1B_1C_1\), two corresponding sides \(AB=A_1B_1\), \(AC=A_1C_1\) and the angles between them \(\angle BAC = \angle B_1A_1C_1\) are equal, then the triangles are congruent.

• If the sides \(AB=A_1B_1\) and pairs of the corresponding angles next to them \(\angle CAB = \angle C_1A_1B_1\) and \(\angle CBA = \angle C_1B_1A_1\) are equal, then the triangles are congruent.

The basic principles about parallel lines and general triangles are:
1. The supplementary angles (angles "hugging" a straight line) add up to \(180^{\circ}\).
2. The sum of all internal angles of a triangle is also \(180^{\circ}\).
3. A line cutting two parallel lines cuts them at the same angles (these are called corresponding angles).
4. In an isosceles triangle (which has two sides of equal lengths), two angles touching the third side are equal.

Is it true that if \(a\) is a positive number, then \(a^2 \ge a\)? What about \(a^2 +1 \ge a\)?

Show for positive \(a\) and \(b\) that \(a^2 +b^2 \ge 2ab\).

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: \[\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b} \ge a +b+c.\]