Problems

Today we will solve some problems using algebraic tricks, mostly related to turning a sum into a product or using an identity involving squares.
The ones particularly useful are: \((a+b)^2 = a^2 +b^2 +2ab\), \((a-b)^2 = a^2 +b^2 -2ab\) and \((a-b) \times (a+b) = a^2 -b^2\). While we are at squares, it is also worth noting that any square of a real number is never a negative number.

The evil warlock found a mathematics exercise book and replaced all the decimal numbers with the letters of the alphabet. The elves in his kingdom only know that different letters correspond to different digits \(\{0,1,2,3,4,5,6,7,8,9\}\) and the same letters correspond to the same digits. Help the elves to restore the exercise book to study.

The perimeter of the triangle \(\triangle ABC\) is \(10\). Let \(D,E,F\) be the midpoints of the segments \(AB,BC,AC\) respectively. What is the perimeter of \(\triangle DEF\)?

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 \(\frac{|AB|}{|BD|}=\frac{|AC|}{|CD|}\).

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 \(a_1\), \(a_2\), ..., \(a_n\): \[\frac{a_1+a_2+...+a_n}{n}\geq \sqrt[n]{a_1a_2...a_n}.\]

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 \[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}.\]

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 \[\frac{1}{a_1} + ... + \frac{1}{a_n} < 1+ \frac{n}{1951}.\]

Suppose you want to compute the sum \(1+2+\dots+n\) up to some positive integer \(n\). Then you discover a curious pattern:

We may guess at this point \(1+2+\dots+n = \frac{n(n+1)}{2}\). How can we prove it? One way is to attack the problem step by step.

When \(n=1\), the sum is just \(1\) and \(\frac{1\times(1+1)}{2}=1\). So far so good.

When \(n=2\), the sum is \(1+2 = 3\) and \(\frac{2\times(2+1)}{2}=3\). We can also see this in another way. As we already noted, \(1 = \frac{1\times(1+1)}{2}\). This means \[1+2 = \frac{1\times (1+1)}{2} + 2 = \frac{1\times2+2\times2}{2} = \frac{(1+2)\times 2}{2} = \frac{2\times(2+1)}{2}.\]

When \(n=3\), we have already proved that \(1 + 2 = \frac{2\times(2+1)}{2}\), so \[1+2+3 = \frac{2\times (2+1)}{2} + 3 = \frac{2\times3+2\times3}{2} = \frac{(2+2)\times 3}{2} = \frac{3\times(3+1)}{2}.\]

When \(n=4\), we have already proved that \(1 + 2 + 3 = \frac{3\times(3+1)}{2}\), so \[1+2+3+4 = \frac{3\times (3+1)}{2} + 4 = \frac{3\times4+2\times4}{2} = \frac{(3+2)\times 4}{2} = \frac{4\times(4+1)}{2}.\]

When \(n=5\), we have already proved that \(1 + 2 + 3 + 4 = \frac{4\times(4+1)}{2}\), so ...

It starts getting a bit boring, but hopefully you get the point. Important takeaways from the example above:

• The truth of the next case depends ONLY on the previous case.

• We know what we need to prove IS true for the first case, that is \(n=1\).

• By repeating the same procedure starting from \(n=1\), we can eventually reach any given positive integer.

• Thus, the formula is true for all positive integers (also known as natural numbers).

A diagram summarizing the idea: \[\text{true for } n=1 \implies \text{true for } n=2 \implies \text{true for } n=3 \implies \text{true for } n=4 \implies \dots\]

This is the mechanism behind induction. Formally, we can state the principle of mathematical induction as follows. Suppose we have a series of statements numbered by the positive integers: 1st statement, 2nd statement, 3rd statement and so on. Suppose that

• the 1st statement is true (the base case is true);

• whenever the \(n\)th statement is true, the \((n+1)\)th statement is also true (the induction step is valid assuming the induction hypothesis).

Then the statement is true for all positive integers (natural numbers). Let us revisit the example and prove it formally now using the principle of mathematical induction.