Problems

Prove the $AM-GM$ inequality for positive real numbers $a_1$, $a_2$, ..., $a_n$: $$\frac{a_1+a_2+...+a_n}{n}\ge \sqrt[n]{a_1{a}_2...a_n}.$$

For non-negative real numbers $a,b,c$ prove that $$a^3+b^3+c^3\ge \frac{(a+b+c)(a^2+b^2+c^2)}{3}\ge a^{2}b+b^{2}c+c^{2}a.$$

Prove Nesbitt’s inequality, which states that for positive real numbers $a,b,c$ we have $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}.$$

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 $1+2+\cdots +n$ up to some positive integer $n$. Then you discover a curious pattern:

We may guess at this point $1+2+\cdots +n=\frac{n(n+1)}{2}$. 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 $$1+2=\frac{1\times (1+1)}{2}+2=\frac{1\times 2+2\times 2}{2}=\frac{(1+2)\times 2}{2}=\frac{2\times (2+1)}{2}.$$

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

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

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: $$\text{true for }n=1⟹\text{true for }n=2⟹\text{true for }n=3⟹\text{true for }n=4⟹\dots$$

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.

Show that if $1+2+\cdots +n=\frac{n(n+1)}{2}$, then $1+2+\cdots +(n+1)=\frac{(n+1)((n+1)+1)}{2}$.

Show that $1+2+\cdots +n=\frac{n(n+1)}{2}$ for every natural number $n$.

Show that if $1+2^1+2^2+\cdots +2^{10}=2^{11}-1$, then $1+2^1+2^2+\cdots +2^{11}=2^{12}-1$.

Show that $1+2^1+2^2+\cdots +2^n=2^{n+1}-1$ 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.