An easy inequality from Nordic mathematical contests !?

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


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?


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.


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}}.


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.


Nalin Pithwa.

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

Reference: Nordic Mathematical Contest 1987, R. Todev:


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.


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


as well as {f(3)}^{8}=f(3^{8})=f(6561)<f(8192)=f(2^{13})={f(2)}^{13}=3^{13}<6^{8}.

So, we arrive at 5<f(3)<6. But, this is not possible, since f(3) is an integer. So, a=4.


Nalin Pithwa.

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

Reference 1:

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

Amazon India link:


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:




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



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.


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.


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.


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


  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