# 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.

# 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.

# Journos have a problem!

A problem posed in Nordic Mathematical Contest 1987:

Question:

Nine journalists, each from a different country, participate in a press conference. None of them can speak more than three languages, and each two journalists have at least one common language. Prove that at least five of the journalists can speak the same language.

Solution1:

Assume the journalists are $J_{1}, J_{2}, J_{3}, \ldots, J_{9}$. Assume that no five of them have a common language. Assume the languages $J_{1}$ speaks are $L_{1}, L_{2}$ and $L_{3}$. Group $J_{2}, J_{3}, \ldots, J_{9}$ according to the language they speak with $J_{1}$. No group can have more than three members. So, either there are three groups of three members each, or two groups with three members and one with two. Consider the first alternative. We may assume that $J_{1}$ speaks $L_{1}$ with $J_{2}$, $J_{3}$, and $J_{4}$, $L_{2}$ with $J_{5}$, $J_{6}$, and $J_{7}$, and $L_{3}$ with $J_{8}$, $J_{9}$ and $J_{2}$. Now, $J_{2}$ speaks $L_{1}$ with $J_{1}$, $J_{3}$, and $J_{4}$, $L_{3}$ with $J_{1}$, $J_{8}$, and $J_{9}$. $J_{2}$ must speak a fourth language, $L_{4}$, with $J_{5}$, $J_{6}$, and $J_{7}$. But, now $J_{5}$ speaks both $L_{2}$ and $L_{4}$ with $J_{2}$, $J_{6}$, and $J_{7}$. So, $J_{5}$ has to use his third language with $J_{1}, J_{4}, J_{8}$ and $J_{9}$. This contradicts the assumption we made.

So, we now may assume that $J_{1}$ speaks $L_{3}$ only with $J_{8}$ and $J_{9}$. As $J_{1}$ is not special, we conclude that for each journalist $J_{k}$, the remaining eight are divided into three mutually exclusive language groups, one of which has only two members. Now, $J_{2}$ uses $L_{1}$ with three others, and there has to be another language he also speaks with three others. If this were $L_{2}$ or $L_{3}$, a group of five would arise (including $J_{1}$). So, $J_{2}$ speaks $L_{4}$ with three among $J_{5}, \ldots, J_{9}$. Either two of these three are among $J_{5}$, $J_{6}$, and $J_{7}$, or among $J_{8}, J_{9}$. Both alternatives lead to a contradiction to the already proved fact that no pair of journalists speaks two languages together. Hence, the proof. QED.

Reference:

Nordic Mathematical Contest, 1987-2009.

https://www.amazon.in/Nordic-Mathematical-Contest-1987-2009-Todev/dp/1450519830/ref=sr_1_1?s=books&ie=UTF8&qid=1515913392&sr=1-1&keywords=Nordic+mathematical+contest

Cheers,

Nalin Pithwa.

# A “primer” on geometric inequalities for pre-RMO and RMO

The comparison of lengths is more basic than comparison of other geometric quantities (such as angles, areas and volumes). A geometric inequality that involves only the lengths is called a distance inequality.

Some simple axioms and theorems on inequalities in Euclidean geometry are usually the starting point to solve the problem of distance inequality, in which most frequently used tools are:

Proposition I:

The shortest line connecting point A with point B is the segment AB.

The direct corollary of proposition 1 is as follows:

Proposition 2:

(Triangle inequality)

For arbitrary three points A, B and C (lying in the same plane), we have $AB \leq AC +CB$, the equality holds iff the three points are collinear.

Remark:

In most literature, any symbol of a geometric object also denotes its quantity according to the context.

Proposition 3:

In a triangle, the longer side has the larger opposite angle. And, conversely, the longer angle has the longer opposite side.

Proposition 4:

The median of a triangle on a side is shorter than the half-sum of the other two sides.

Proposition 5:

If a convex polygon is within another one, then the outside convex polygon’s perimeter is larger.

Proposition 6:

Any segment in a convex polygon is either less than the longest side or the longest diagonal of the convex polygon.

Here, is a classic example:

Example 1:

Let $a, b, c$ be the sides of a $\triangle ABC$. Prove that $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}<2$.

Solution 1:

By the triangle inequality, $a yields $\frac{a}{b+c} = \frac{2a}{2(b+c)} < \frac{2a}{a+b+c}$

Similarly, $\frac{b}{c+a} < \frac{2b}{a+b+c}$ and $\frac{c}{b+a} < \frac{2c}{a+b+c}$

Adding up the above three inequalities, we get the required inequality.

Example 2:

Let AB be the longest side of $\triangle ABC$, and P a point in the triangle, prove that $PA+PB>PC$.

Solution/Proof 2:

Let D be the intersection point of CP and AB. (Note P is in the interior of the triangle ABC). Then, $\angle ADC$ or $\angle BDC$ is not acute. Without loss of generality, we assume that $\angle ADC$ is not acute. Applying proposition 3 to $\triangle ADC$, we obtain $AC \geq CD$. Therefore, $AB \geq AC \geq CD > PC$….call this as  relationship “a”.

Furthermore, applying triangle inequality to $\triangle PAB$, we have $PA+PB>AB$…call this as relationship “b”.

Combining “a” and “b”, we obtain the required inequality immediately. QED.

Remarks: (1) If AB is not the longest, then the conclusion may not be true. (2) If point P on the plane of regular triangle ABC, P is not on the circumcircle of the triangle, then the sum of any two of PA, PB, and PC is longer than the remaining one. That is, PA, PB and PC consist of a triangle’s three sides.

Quiz:

Prove that a closed polygonal line with perimeter 1 can be put inside a circle with radius 0.25.

Reference:

Geometric Inequalities, Vol 12, Gangsong Leng, translated by Yongming Liu.

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.

# A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra

Problem:

Let a and b satisfy $a \geq b >0$ and $a+b=1$.

1. Prove that if m and n are positive integers with $m, then $a^{m}-a^{n} \geq b^{m}-b^{n}>0$.
2. For each positive integer n, consider a quadratic function: $f_{n}(x)=x^{2}-b^{n}x-a^{n}$.

Show that $f(x)$ has two roots that are in between -1 and 1.

Solution:

Let $k=n-m \in (0,n)$. Consider $a^{m}b-ab^{m}=ab(a^{m-1}-b^{m-1})$ with $m-1 \geq 0$. Since $a \geq b >0$, we have $a^{m-1} \geq b^{m-1}$. Hence, $a^{m}b \geq ab^{m}$. Call this relationship I.

On the other hand, notice that $a \geq b$, $a^{2} \geq b^{2}$, $\ldots$, $a^{k-1} \geq b^{k-1}$, which implies

$1+a+a^{2}+\ldots+a^{k-1} \geq 1+b+b^{2}+ \ldots + b^{k-1}$….call this relationship II.

From relationships I and II, it follows that

$a^{m}b(1+a+a^{2}+\ldots+a^{k-1}) \geq ab^{m}(1+b+b^{2}+\ldots+b^{k-1})$

which can be written as

$a^{m}(1-a)(1+a+a^{2}+\ldots+a^{k-1}) \geq b^{m}(1-b)(1+b+b^{2}+\ldots+b^{k-1})$, or equivalently, $a^{m}(1-a^{k}) \geq b^{m}(1-b^{k})$. That is, $a^{m}-a^{n} \geq b^{m}-b^{n}$.

It remains to prove that $b^{m}-b^{n}>0$. Indeed, $b^{m}-b^{n}=b^{m}(1-b^{k})>0$ as $0.

The equality occurs if and only if $a=b=\frac{1}{2}$.

2) Since discriminant $\Delta=b^{2n}+4a^{n}>0$, $f_{n}(x)$ has two distinct real roots $x_{1} \neq x_{2}$. Also, note that if $a, b \in (0,1)$, then the following holds:

$f_{n}(1)=1-b^{n}-a^{n}=a+b-b^{n}-a^{n}=(a-a^{n})+(b-b^{n}) \geq 0$,

$f_{n}(-1)=1+b^{n}-a^{n}=(1-a^{n})+b^{n} \geq 0$,

$\frac{S}{2} = \frac{x_{1}+x_{2}}{2}=\frac{b^{n}}{2} \in (-1,1)$.

We conclude that $x_{1}, x_{2} \in [-1, 1]$.

Cheers,

Nalin Pithwa.

Reference: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, Le Hai Khoi.