# How to find the number of proper divisors of an integer and other cute related questions

Question 1:

Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):

Solution 1:

Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, $441000 = (2^{3})(3^{2})(5^{3})(7^{2})$. Any divisor, proper or improper, of the given number must be of the form $(2^{a})(3^{b})(5^{c})(7^{d})$ where $0 \leq a \leq 3$, $0 \leq b \leq 2$, $0 \leq c \leq 3$, and $0 \leq d \leq 2$. In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be $(4)(3)(4)(3)-2=142$.

Question 2:

Count the proper divisors of an integer N whose prime factorization is: $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}$

Solution 2:

By using the same reasoning as in previous question, the number of proper divisors of N is $(\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2$, where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.

Question 3:

Find the number of ways of factoring 441000 into 2 factors, m and n, such that $m>1, n>1$, and the GCD of m and n is 1.

Solution 3:

Consider the set $A = {2^{3}, 3^{2}, 5^{3}, 7^{2}}$ associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).

The possible unordered partitions are the following:

$A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\}$,

and $A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}$

Hence, the required answer is $4+3=2^{4-1}+1=7$.

Question 4:

Generalize the above problem by showing that any integer has $2^{k-1}-1$ factorizations into relatively prime pairs m, n ($m>1, n>1$).

Solution 4:

Proof by mathematical induction on k:

For $k=1$, the result holds trivially.

For $k=2$, we must prove that a set of k distinct elements, $Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\}$ has $2^{k-1}-1$ sets. Now, one partition of Z is

$Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W$

All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in $2^{k-2}-1$ ways — and then, including $a_{k}$ in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore

$1 + (2^{k-2})(2)=2^{k-1}-1$. QED.

Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case.

Cheers,

Nalin Pithwa.

# Some basics of Number Theory for RMO: part III: Fermat’s Little Theorem

Fermat’s Little Theorem:

The fact that there are only a finite number of essentially different numbers in arithmetic to a modulus m means that there are algebraic relations which are satisfied by every number in that arithmetic. There is nothing analogous to these relations in ordinary arithmetic.

Suppose we take any number x and consider its powers $x, x^{2}, x^{3}, \ldots$. Since there are only a finite number of possibilities of these to the modulus m, we must eventually come to one which we have met before, say $x^{h} \equiv x^{k} {\pmod m}$, where $k . If x is relatively prime to m, the factor $x^{k}$ can be cancelled, and it follows that $x^{l} \equiv 1 {\pmod m}$, where $l \equiv {h-k}$. Hence, every number x which is relatively prime to m satisfies some congruence of this form. The least exponent l for which $x^{l} \equiv 1 {\pmod m}$ will be called the order of x to the modulus m. If x is 1, its order is obviously 1. To illustrate the definition, let us calculate the orders of a few numbers to the modulus 11. The powers of 2, taken to the modulus 11, are

2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, $\ldots$

Each one is twice the preceding one, with 11 or a multiple of 11 subtracted where necessary to make the result less than 11. The first power of 2 which is $\equiv 1$ is $2^{10}$, and so the order of $2 \pmod {11}$ is 10. As another example, take the powers of 3:

3, 9, 5, 4, 1, 3, 9, $\ldots$

The first power of 3 which is equivalent to 1 is $3^{5}$, so the order of $3 \pmod {11}$ is 5. It will be found that the order of 4 is again 5, and so also is that of 5.

It will be seen that the successive powers of x are periodic; when we have reached the first number l for which $x^{l} \equiv 1$, then $x^{l+1} \equiv x$ and the previous cycle is repeated. It is plain that $x^{n} \equiv 1 {\pmod m}$ if and only if n is a multiple of the order of x. In the last example, $3^{n} \equiv 1 {\pmod 11}$ if and only if n is a multiple of 5. This remains valid if n is 0 (since 3^{0} = 1), and it remains valid also for negative exponents, provided $3^{-n}$, is interpreted as a fraction (mod 11) in the way explained earlier (an earlier blog article).

In fact, the negative powers of 3 (mod 11) are obtained by prolonging the series backwards, and the table of powers of 3 to the modulus 11 is:

$\begin{array}{cccccccccccccc} n & = & \ldots & -3 & -2 & -1 & 0 & 1 &2 & 3 & 4 & 5 & 6 & \ldots \\ 3^{n} & \equiv & \ldots & 9 & 5 & 4 & 1 & 3 & 9 & 5 & 4 & 1 & 3 & \ldots \end{array}$

Fermat discovered that if the modulus is a prime, say p, then every integer x not congruent to 0 satisfies

$x^{p-1} \equiv 1 {\pmod p}$….call this as equation A.

In view of what we have seen above, this is equivalent to saying that the order of any number is a divisor of $p-1$. The result A was mentioned by Fermat in a letter to Frenicle de Bessy of 18 October 1640, in which he also stated that he had a proof. But, as with most of Fermat’s discoveries, the proof was not published or preserved. The first known proof seems to have been given by Leibniz (1646-1716). He proved that $x^{p} \equiv x {\pmod p}$, which is equivalent to A, by writing x as a sum $1+ 1 + 1 + \ldots + 1$ of x units (assuming x positive), and then expanding $(1+1+ \ldots + 1)^{p}$ by the multinomial theorem. The terms $1^{p} + 1^{p} + \ldots + 1^{p}$ give x, and the coefficients of all the other terms are easily proved to be divisible by p.

Quite a different proof was given by Ivory in 1806. If $x \not\equiv 0 {\pmod p}$, the integers

$x, 2x, 3x, \ldots, (p-1)x$

are congruent in some order to the numbers

$1, 2, 3, \ldots, p-1$.

In fact, each of these sets constitutes a complete set of residues except that 0 (zero) has been omitted from each. Since the two sets are congruent, their products are congruent, and so

$(x)(2x)(3x) \ldots ((p-1)x) \equiv (1)(2)(3)\ldots (p-1){(\pmod p)}$

Cancelling the factors 2, 3, ….(p-1), as is permissible we obtain the above relation A.

One merit of this proof is that it can be extended so as to apply to the more general case when the modulus is no longer a prime.

The generalization of the result A to any modulus was first given by Euler in 1760. To formulate it, we must begin by considering how many numbers in the set 0, 1, 2, …, (m-1) are relatively prime to m. Denote this number by $\phi(m)$. When m is a prime, all the numbers in the set except 0 (zero) are relatively prime to m, so that $\phi(p) = p-1$ for any prime p. Euler’s generalization of Fermat’s theorem is that for any modulus m,

$x^{\phi(m)} = 1 {\pmod m}$…relation B

provided only that x is relatively prime to m.

To prove this, it is only necessary to modify Ivory’s method by omitting from the numbers $0, 1, 2, \ldots, (m-1)$ not only the number 0, but all numbers which are not relatively prime to m. These remain $\phi(m)$ numbers, say

$a_{1}, a_{2}, \ldots, a_{\mu}$, where $\mu = \phi(m)$.

Then, the numbers

$a_{1}x, a_{2}x, \ldots, a_{\mu}x$

are congruent, in some order, to the previous numbers, and on multiplying and cancelling $a_{1}, a_{2}, \ldots, a_{\mu}$ (as is permissible) we obtain $x^{p} \equiv 1 {\pmod m}$, which is relation B.

To illustrate this proof, take $m=20$. The numbers less than 20 and relatively prime to 20 are :

1, 3, 7, 9, 11, 13, 17, 19.

So that $\phi(20) = 8$. If we multiply these by any number x which is relatively prime to 20, the new numbers are congruent to the original numbers in some other order. For example, if x is 3, the new numbers are congruent respectively to

$3, 9, 1, 7, 13, 19, 11, 17 {\pmod 20}$;

and the argument proves that $3^{8} \equiv 6561$.

Reference:

1. The Higher Arithmetic, H. Davenport, Eighth Edition.
2. Elementary Number Theory, Burton, Sixth Edition.
3. A Friendly Introduction to Number Theory, J. Silverman

Shared for those readers who enjoy expository articles.

Nalin Pithwa.

# Combinatorics for RMO : some basics and examples: homogeneous products of r dimensions

Question:

Find the number of homogeneous products of r dimensions that can be formed out of the n letters a, b, c ….and their powers.

Solution:

By division, or by the binomial theorem, we have:

$\frac{1}{1-ax} = 1 + ax + a^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-bx} = 1+ bx + b^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-cx} = 1 + cx + c^{2}x^{2} + c^{3}x^{3} + \ldots$

Hence, by multiplication,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \times \ldots$

$= (1+ax + a^{2}x^{2}+a^{3}x^{3}+ \ldots)(1+bx + b^{2}x^{2} + b^{3}x^{3}+ \ldots)(1+cx + c^{2}x^{2} + c^{3}x^{3}+ \ldots)\ldots$

$= 1 + x(a + b + c + \ldots) +x^{2}(a^{2}+ab+ac+b^{2}+bc + c^{2} + \ldots) + \ldots$

$= 1 + S_{1}x + S_{2}x^{2} + S_{3}x^{3} + \ldots$ suppose;

where $S_{1}$, $S_{2}$, $S_{3}$, $\ldots$ are the sums of the homogeneous products of one, two, three, … dimensions that can be formed of a, b, c, …and their powers.

To obtain the number of these products, put a, b, c, …each equal to 1; each term in $S_{1}$, $S_{2}$, $S_{3}$, …now becomes 1, and the values of $S_{1}$, $S_{2}$, $S_{3}$, …so obtained give the number of the homogeneous products of one, two, three, ….dimensions.

Also,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \ldots$

becomes $\frac{1}{(1-x)^{n}}$, or $(1-x)^{-n}$

Hence, $S_{r} =$ the coefficient of $x^{r}$ in the expansion of $(1-x)^{-n}$

$= \frac{n(n+1)(n+2)(n+3)\ldots (n+r-1)}{r!}= \frac{(n+r-1)!}{r!(n-1)!}$

Question:

Find the number of terms in the expansion of any multinomial when the index is a positive integer.

In the expansion of $(a_{1}+ a_{2} + a_{3} + \ldots + a_{r})^{n}$

every term is of n dimensions; therefore, the number of terms is the same as the number of homogeneous products of n dimensions that can be formed out of the r quantities $a_{1}$, $a_{2}$, $a_{3}$, …$a_{r}$, and their powers; and therefore by the preceding question and solution, this is equal to

$\frac{(r+n-1)!}{n! (r-1)!}$

A theorem in combinatorics:

From the previous discussion in this blog article, we can deduce a theorem relating to the number of combinations of n things.

Consider n letters a, b, c, d, ….; then, if we were to write down all the homogeneous products of r dimensions, which can be formed of these letters and their powers, every such product would represent one of the combinations, r at a time, of the n letters, when any one of the letters might occur once, twice, thrice, …up to r times.

Therefore, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of homogeneous products of r dimensions which can be formed out of n letters, and therefore equal to $\frac{(n+r-1)!}{r!(n-1)!}$, or ${{n+r-1} \choose r}$.

That is, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of combinations of $n+r-1$ things r at a time when repetitions are NOT allowed.

Example 1:

Find the coefficient of $x^{r}$ in the expansion of $\frac{(1-2x)^{2}}{(1+x)^{3}}$

Solution 1:

The expression $= (1-4x+4x^{2})(1+p_{1}x+p_{2}x^{2}+ \ldots + p_{r}x^{r}+ \ldots)$, suppose.

The coefficients of $x^{r}$ will be obtained by multiplying $p_{r}$, $p_{r-1}$, $p_{r-2}$ by 1, -4, and 4 respectively, and adding the results; hence,

the required coefficient is $p_{r} - 4p_{r-1}+4p_{r-2}$

But, with a little work, we can show that $p_{r} = (-1)^{r}\frac{(r+1)(r+2)}{2}$.

Hence, the required coefficient is

$= (-1)^{r}\frac{(r+1)(r+2)}{2} - 4(-1)^{r-1}\frac{r(r+1)}{2} + 4 (-1)^{r-2}\frac{r(r-1)}{2}$

$= \frac{(-1)^{r}}{2}\times ((r+1)(r+2) + 4r(r+1) + 4r(r-1))$

$= \frac{(-1)^{r}}{2}(9r^{2}+3r+2)$

Example 2:

Find the value of the series

$2 + \frac{5}{(2!).3} + \frac{5.7}{3^{2}.(3!)} + \frac{5.7.9}{3^{3}.(4!)} + \ldots$

Solution 2:

The expression is equal to

$2 + \frac{3.5}{2!}\times \frac{1}{3^{2}} + \frac{3.5.7}{3!}\times \frac{1}{3^{3}} + \frac{3.5.7.9}{4!}\times \frac{1}{3^{4}} + \ldots$

$= 2 + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times \frac{2^{2}}{3^{2}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times \frac{2^{3}}{3^{3}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times \frac{2^{4}}{3^{4}} + \ldots$

$= 1 + \frac{\frac{3}{2}}{1} \times \frac{2}{3} + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times (\frac{2}{3})^{2} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times (\frac{2}{3})^{3} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times (\frac{2}{3})^{4} + \ldots$

$= (1-\frac{2}{3})^{\frac{-3}{2}} = (\frac{1}{3})^{-\frac{3}{2}} = 3^{\frac{3}{2}} = 3 \sqrt{3}$.

Example 3:

If n is any positive integer, show that the integral part of $(3+\sqrt{7})^{n}$ is an odd number.

Solution 3:

Suppose I to denote the integral and f the fractional part of $(3+\sqrt{7})^{n}$.

Then, $I + f = 3^{n} + {n \choose 1}3^{n-1}\sqrt{7} + {n \choose 2}3^{n-2}.7 + {n \choose 3}3^{n-3}.(\sqrt{7})^{3}+ \ldots$…call this relation 1.

Now, $3 - \sqrt{7}$ is positive and less than 1, therefore $(3-\sqrt{7})^{n}$ is a proper fraction; denote it by $f^{'}$;

Hence, $f^{'} = 3^{n} - {n \choose 1}.3^{n-1}.\sqrt{7} + {n \choose 2}.3^{n-2}.7 - {n \choose 3}.3^{n-3}.(\sqrt{7})^{3}+ \ldots$…call this as relation 2.

Add together relations 1 and 2; the irrational terms disappear, and we have

$I + f + f^{'} = 2(3^{n} + {n \choose 2}.3^{n-2}.7+ \ldots ) = an even integer$

But, since f and $f^{'}$ are proper fractions their sum must be 1;

Hence, I is an odd integer.

Nalin Pithwa.

# Pre-RMO or RMO algebra practice problem: infinite product

Find the product of the following infinite number of terms:

$\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \ldots = \prod_{m=2}^{\infty}\frac{m^{3}-1}{m^{3}+1}$

$m^{3}-1=(m-1)(m^{2}+m+1)$, and also, $m^{3}+1=(m+1)(m^{2}-m+1)=(m-1+2)((m-1)^{2}+(m-1)+1)$

Hence, we get $P_{m}=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \ldots \times \frac{m^{3}-1}{m^{3}+1}$, which in turn, equals

$(\frac{1}{3} \times \frac{7}{3}) \times (\frac{2}{4} \times \frac{13}{7}) \times (\frac{3}{5} \times \frac{21}{13})\times \ldots (\frac{m-1}{m+1} \times \frac{m^{2}+m+1}{m^{2}-m+1})$, that is, in turn equal to

$\frac{2}{3} \times \frac{m^{2}+m+1}{m(m+1)}$, that is, in turn equal to

$\frac{}{} \times (1+ \frac{1}{m(m+1)})$, so that when $m \rightarrow \infty$, and then $P_{m} \rightarrow 2/3$.

personal comment: I did not find this solution within my imagination !!! 🙂 🙂 🙂

The credit for the solution goes to “Popular Problems and Puzzles in Mathematics” by Asok Kumar Mallik, IISc Press, Foundation Books. Thanks Prof. Mallik !!

Cheers,

Nalin Pithwa

# Algebra Training for RMO and Pre-RMO: let’s continue: Fibonacci Problem and solution

Fibonacci Problem:

Leonardo of Pisa (famous as Fibonacci) (1173) wrote a book “Liber Abaci” (1202), wherein he introduced Hindu-Arabic numerals in Europe. In 1225, Frederick II declared him as the greatest mathematician in Europe when he posed the following problem to defeat his opponents.

Question:

Determine the rational numbers x, y and z to satisfy the following equations:

$x^{2}+5=y^{2}$ and $x^{2}-5=z^{2}$.

Solution:

Definition: Euler defined a congruent number to be a rational number that is the area of a right-angled triangle, which has rational sides. With p, q, and r as a Pythagorean triplet such that $r^{2}=p^{2}+q^{2}$, then $\frac{pq}{2}$ is a congruent number.

It can be shown that square of a rational number cannot be a congruent number. In other words, there is no right-angled triangle with rational sides, which has an area as 1, or 4, or $\frac{1}{4}$, and so on.

Characteristics of a congruent number: A positive rational number n is a congruent number, if and only if there exists a rational number u such that $u^{2}-n$ and $u^{2}+n$ are the squares of rational numbers. (Thus, the puzzle will be solved if we can show that 5 is a congruent number and we can determine the rational number $u(=x)$). First, let us prove the characterisitic mentioned above.

Necessity: Suppose n is a congruent number. Then, for some rational number p, q, and r, we have $r^{2}=p^{2}+q^{2}$ and $\frac{pq}{2}=n$. In that case,

$\frac{p+q}{2}, \frac{p-q}{2}$ and n are rational numbers and we have

$(\frac{p+q}{2})^{2}=\frac{p^{2}+q^{2}}{4}+\frac{pq}{2}=(\frac{r}{2})^{2}+n$ and similarly,

$(\frac{p-q}{2})^{2}=(\frac{r}{2})^{2}-n$.

Setting $u=\frac{r}{2}$, we get $u^{2}-n$ and $u^{2}+n$ are squares of rational numbers.

Sufficiency:

Suppose n and u are rational numbers such that $\sqrt{u^{2}-n}$ and $\sqrt{u^{2}+n}$ are rational, when

$p=\sqrt{u^{2}+n}+\sqrt{u^{2}-n}$ and $q=\sqrt{u^{2}+n}-\sqrt{u^{2}-n}$

and 2n are rational numbers satisfying $p^{2}+q^{2}=(2n)^{2}$ is a rational square and also $\frac{pq}{2}=n$, a rational number which is a congruent number.

So, we see that the Pythagorean triplets can lead our search for a congruent number. Sometimes a Pythagorean triplet can lead to more than one congruent number as can be seen with $(9,40,41)$. This set obviously gives 180 as a congruent number. But, as $180=5 \times 36=5 \times 6^{2}$, we can also consider a rational Pythagorean triplet $(\frac{9}{6}, \frac{40}{6}, \frac{41}{6})$, which gives a congruent number 5 (we were searching for this congruent number in this puzzle!). We also determine the corresponding $u=\frac{r}{2}=\frac{41}{12}$.

The puzzle/problem is now solved with $x=\frac{41}{12}$, which gives $y=\frac{49}{12}$, and $z=\frac{31}{12}$.

One can further show that if we take three rational squares in AP, $u^{2}-n$, and $u^{2}$, and $u^{2}+n$, with their product defined as a rational square $v^{2}$ and n as a congruent number, then $x=u^{2}$, $y=v$ is a rational point on the elliptic curve $y^{2}=x^{3}-n^{2}x$.

Reference:

1) Popular Problems and Puzzles in Mathematics: Asok Kumar Mallik, IISc Press, Foundation Books, Amazon India link:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1525099935&sr=1-1&keywords=popular+problems+and+puzzles+in+mathematics

2) Use the internet, or just Wikipedia to explore more information on Fibonacci Numbers, Golden Section, Golden Angle, Golden Rectangle and Golden spiral. You will be overjoyed to find relationships amongst all the mentioned “stuff”.

Cheers,

Nalin Pithwa

# An easy inequality from Nordic mathematical contests !?

Reference: Nordic Mathematical Contest, 1987-2009, R. Todev.

Question:

Let a, b, and c be real numbers different from 0  and $a \geq b \geq c$. Prove that inequality

$\frac{a^{3}-c^{3}}{3} \geq abc(\frac{a-b}{c} + \frac{b-c}{a})$

holds. When does the equality hold?

Proof:

We know that a, b and c are real, distinct and also non-zero and also that $a \geq b \geq c$.

Hence, $c-b \leq 0 \leq a-b$, we have $(a-b)^{3}\geq (c-b)^{3}$, or

$a^{3}-3a^{a}b+3ab^{2}-b^{3} \geq c^{3}-3bc^{2}+3b^{2}c-b^{3}$

On simplifying this, we immediately have

$\frac{1}{3}{(a^{3}-c^{3})} \geq a^{2}b-ab^{2}+b^{2}c-bc^{2}=abc(\frac{a-b}{c}+\frac{b-c}{a})$.

A sufficient condition for equality is $a=c$. If $a>c$, then $(a-b)^{3}>(c-b)^{3}$. which makes the proved inequality a strict one. So, $a=c$ is a necessary condition for equality too.

-Nalin Pithwa.

# There are many “inequalities” ! :-( :-) !

Reference: R. Todev, Nordic Mathematical Contests, 1987-2009.

Question:

Let a, b, and c be positive real numbers. Prove that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \leq \frac{a^{2}}{b^{2}} + \frac{b^{2}}{c^{2}} + \frac{c^{2}}{a^{2}}$.

Solution:

The arithmetic-geometric inequality yields

$3=3\sqrt[3]{\frac{a^{2}}{b^{2}}.\frac{b^{2}}{c^{2}}.\frac{c^{2}}{a^{2}}}\leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}$,

or $\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}} + \frac{b^{2}}{c^{2}} + \frac{c^{2}}{a^{2}}}$…call this relation I.

On the other hand, the Cauchy-Schwarz inequality implies

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \leq \sqrt{1^{2}+1^{2}+1^{2}}\sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}=\sqrt{3}\sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}$….call this relation II.

We arrive at the inequality we desire by combining relations I and II. Hence, the proof. QED.

Cheers,

Nalin Pithwa.