# 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.

# Functions — “s’wat” Math is about !! :-)

Reference: Nordic Mathematical Contest 1987, R. Todev:

Question:

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions $f(2)=a>2$ and $f(mn)=f(m)f(n)$ for all natural numbers m and n. Define the smallest possible value of a.

Solution:

Since, $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that $a=3$. It is easy to prove by induction that $f(n^{k})={f(n)}^{k}$ for all $k \geq 1$. So, taking into account that f is strictly increasing, we get

${f(3)}^{4}=f(3^{4})=f(81)>f(64)=f(2^{6})={f(2)}^{6}=3^{6}=27^{2}>25^{2}=5^{4}$

as well as ${f(3)}^{8}=f(3^{8})=f(6561).

So, we arrive at $5. But, this is not possible, since $f(3)$ is an integer. So, $a=4$.

Cheers,

Nalin Pithwa.

# Francois Viete, France, Vietnam and Algebra: RMO training for algebra

Reference 1: https://en.wikipedia.org/wiki/Vieta%27s_formulas

Reference 2: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, and Le Hai Khoi, published by World Scientific;

Question:

Without solving the cubic equation, $x^{3}-x+1=0$, compute the sum of the eighth powers of all roots of the equation.

Approach: we want to be able to express the sum of the eighth powers of the three roots in terms of the three Viete’s relations here.

If $x_{1}$, $x_{2}$, $x_{3}$ are roots of the given cubic equation then, by Viete’s relations between roots and coefficients, we can say the following:

$x_{1}+x_{2}+x_{3}=0$

$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}=-1$

$x_{1}x_{2}x_{3}=-1$.

Furthermore, from $x_{i}^{3}-x_{i}+1=0$, it follows that

$x_{i}^{3}=x_{i}-1$

$x_{i}^{5}=x_{i}^{3}.x_{i}^{2}=(x_{i}-1)x_{i}^{2}=x_{i}^{3}-x_{i}^{2}=-x_{i}^{2}+x_{i}-1$

Then, $x_{1}^{8}+x_{2}^{8}+x_{3}^{8}=2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}) - 3(x_{1}+x_{2}+x_{3})+6$

But, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{2}-2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=2$ and so $x_{1}^{8}+x_{2}^{8}+x_{3}^{8}=4-0+6=10$.

More later,

Nalin Pithwa.

# An inequality for harmonic numbers — RMO Inequalities — Basics

The Harmonic Numbers $H_{j}$ for $j=1,2,3, \ldots$ are defined by

$H_{j}=1+\frac{1}{2}+\frac{1}{3} +\ldots + \frac{1}{j}$

For instance, $H_{4}=1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4} = \frac{25}{12}$

Use mathematical induction to show that $H_{2^{n}} \geq 1 + \frac{n}{2}$ whenever n is a non-negative integer.

Solution:

To carry out the proof, let P(n) be the proposition that $H_{2^{n}}=1+\frac{n}{2}$

Basis step:

P(0) is true because $H_{2^{0}}=H_{1}=1 \geq 1 + \frac{0}{2}$

Inductive Step:

The inductive hypothesis is the statement that $P(k)$ is true, that is, $H_{2^{k}} \geq 1 + \frac{k}{2}$

where k is a non-negative integer. We must show that if $P(k)$ is true, then $P(k+1)$, which states that

$H_{2^{k+1}} \geq 1 + \frac{k+1}{2}$, is also true. So, assuming the inductive hypothesis, it follows that

$H_{2^{k+1}} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^{k}} + \frac {1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$…this step follows from the definition of harmonic number

$=H_{2^{k}} + \frac{1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$….this step again follows by the definition of $2^{k}$th harmonic number

$\geq (1+\frac{k}{2}) + \frac{1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$…this step follows by the inductive hypothesis

$\geq (1+ \frac{k}{2}) + 2^{k}. \frac{2}{2^{k+1}}$…because there are $2^{k}$ terms each greater than or equal to $\frac{1}{2^{k+1}}$

$\geq (1+\frac{k}{2})+ \frac{1}{2}$….canceling a common factor of $2^{k}$ in second term

$= 1 + \frac{k+1}{2}$

This establishes the inductive step of the proof.

We have completed the basis step and the inductive step. Thus, by mathematical induction $P(n)$ is true for all non-negative integers. That is, the inequality $H_{2^{n}} \geq 1 + \frac{n}{2}$ for the harmonic numbers is valid for all non-negative integers n. QED.

Remark:

The inequality established here shows that the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots$ is a divergent series. This is an important example in the study of infinite series.

Note:

Google, or through some other literature, find out why harmonic numbers are termed so. You will understand something more beautiful, more deeper !!

Nalin Pithwa.

# Recurrence Relations problem set 1 — RMO training

Exercises:

1. If $0 \le k \le \lfloor \frac{n}{2} \rfloor$, show that ${n \choose k} \le {n \choose {k+1}}$
2. Prove by induction on n that $\lfloor n \rfloor$ has $2^{n}$ subsets.
3. Show that $1.1! + 2.2! + 3.3! + \ldots + n.n! = (n+1)!-1$
4. Show that ${n \choose k}{k \choose l} = {n \choose l}{{n-l} \choose {k-l}}$ for each $n \ge k \ge l \ge 0$
5. Show that ${n \choose k} = {{n-1} \choose k} + {{n-1} \choose {k-1}}$
6. Show that ${{n+k+1} \choose {k}} = \sum_{i=0}^{n} {{n+i} \choose {i}}$
7. Show that ${{m+n} \choose {k}} = \sum_{i=0}^{k}{m \choose i}{{n} \choose {k-i}}$
8. Show that $\frac{2^{2n}}{2n+1} < {{2n} \choose {n}} < 2^{2n}$ and use Stirling’s formula to prove that ${{2n} \choose {n}} \sim \frac{2^{2n}}{\sqrt{\pi n}}$
9. Give a solution using binomial coefficients and a direct combinatorial solution to the following question: How many pairs $(A,B)$ of subsets of $[n]$ are there such that $A \bigcap B = \phi$?
10. Show that the number of even subsets of $[n]$ equals the number of odd subsets of $[n]$. Give two proofs, one using binomial formula, and one using a direct bijection. Calculate the sum of the sizes of all even (odd) subsets of $[n]$.

More later,

Nalin Pithwa

# Heights and Distances — II — problems for RMO and IITJEE maths

This is one of the prime utilities of trigonometry —- to calculate heights and distances, called as surveying in civil engineering.

1. From the extremities of a horizontal base line AB, whose length is 1 km, the bearings of the foot C of a tower are observed and it is found that $\angle {CAB}$ is 56 degrees and 23 minutes and $\angle{CBA}$ is 47 degrees and 15 minutes, and the elevation of the tower from A is 9 degrees and 25 minutes, find the height of the tower.

2. A man in a balloon observes that the angle of depression of an object on the ground bearing due north is 33 degrees; the balloon drifts 3 km due west and the angle of depression is now found to be 21 degrees. Find the height of the balloon.

3. A tower PN stands on level ground. A base AB is measured at right angles to AN, the points A, B, and N being in the same horizontal plane, and the angles PAN and PBN are found to be $\alpha$ and $\beta$ respectively. Prove that the height of the tower is

AB$\frac{\sin {\alpha}\sin{\beta}}{\sqrt{\sin{(\alpha-\beta)}\sin{(\alpha+\beta)}}}$

If AB is 100m, $\alpha = 70$ degrees and $\beta = 50$ degrees, calculate the height.

4. At each end of a horizontal base of length 2a, it is found that the angular height of a certain peak is $\theta$ and that at the middle point it is $\phi$. Prove that the vertical height of the peak is

$\frac{a \sin {\theta}\sin{\phi}}{\sqrt{\sin{(\phi+\theta)}\sin{(\phi-\theta)}}}$

5. To find the distance from A to P, a distance AB of 1 km. is measured for a convenient direction. At A the angle PAB is found to be 41 degrees 18 min and at B, the angle PBA is found to be 114 degrees and 38 min. What is the required distance to the nearest metre?

More later,

Nalin Pithwa

# Heights and Distances — I — problems for RMO and IITJEE Maths

1. A man observes that at a point due south of a certain tower its angle of elevation is 60 degrees. He then walks 100 meters due west on a horizontal plane and finds that the angle of elevation is 30 degrees. Find the height of the tower and his original distance from it.

2. At the foot of a mountain the elevation of its summit is found to be 45 degrees; after ascending 1000 m towards the mountain up a slope of 30 degree inclination, the elevation is found to be 60 degrees. Find the height of the mountain.

3. A square tower stands upon a horizontal plane, from which three of its upper corners are visible, their angular elevations are respectively 45 degrees, 60 degrees and 45 degrees. Show  that the height of the tower is to the breadth of one of its sides as $\sqrt{6}(\sqrt{5}+1)$ to 4.

4. A lighthouse, facing north, sends out a fan-shaped beam of light extending from north-east to north-west. An observer on a steamer, sailing due west, first sees the light when is 5 km, away from the lighthouse and continues to see it for $30\sqrt{2}$ minutes. What is the speed of the steamer?

5. A man stands at a point X on the bank XY of a river with straight and parallel banks, and observes that the line joining X to a point Z on the opposite bank makes an angle of 30 degrees with XY. He then goes along the bank a distance of 200 meters to Y and finds that the angle ZYX is 60 degrees. Find the breadth of the river.

6. A man, walking due north, observes that the elevation of a balloon, which is due east of him and is sailing toward the north-west, is then 60 degrees; after he has walked 400 meters the balloon is vertically over his head; find its height supposing it to have always remained the same.

The above are some tricky trig problems ! What is the main trick ? I will suggest a very simple hint: draw as good diagrams as possible!

Nalin Pithwa