Solutions to two algebra problems for RMO practice

Problem 1.

If a, b, c are non-negative real numbers such that (1+a)(1+b)(1+c)=8, then prove that the product abc cannot exceed 1.

Solution I:

Given that a \geq 0, b \geq 0, c \geq 0, so certainly abc>0, ab>0, bc>0, and ac>0.

Now, (1+a)(1+b) = 1 + a + b + ab and hence, (1+a)(1+b)(1+c) = (1+a+b+ab)(1+c)= 1+a+b+ab+c +ac + bc + abc=8, hence we get:

a+b+c+ab+bc+ca+abc=7.Β Clearly, the presence ofΒ a+b+c and abc reminds us of the AM-GM inequality.

Here it is AM \geq GM.

So, \frac{a+b+c}{3} \geq (abc)^{1/3}.

Also, we can say: \frac{ab+bc+ca}{3} \geq (^{1/3}. Now, let x=(abc)^{1/3}.

So, 8 \geq 1+3x+3x^{2}+x^{3}

that is, 8 \geq (1+x)^{3}, or 2 \geq 1+x, that is, x \leq 1.Β So, this is a beautiful application of arithmetic mean-geometric mean inequality twice. πŸ™‚ πŸ™‚

Problem 2:

If a, b, c are three rational numbers, then prove that :\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}} is always the square of a rational number.

Solution 2:

Let x=\frac{1}{a-b}, y=\frac{1}{b-c}, z=\frac{1}{c-a}. It can be very easily shown that \frac{1}{x}+ \frac{1}{y} + \frac{1}{z} =0, or xy+yz+zx=0. So, the given expression x^{2}+y^{2}+z^{2}=(x+y+z)^{2} is a perfect square !!!Β BINGO! πŸ™‚ πŸ™‚ πŸ™‚

Nalin Pithwa.

Euler Series question and solution


Mengoli had posed the following series to be evaluated:

1+ \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \ldots.

Some great mathematicians, including Liebnitz, John Bernoulli and D’Alembert, failed to compute this infinite series. Euler established himself as the best mathematician of Europe (in fact, one of the greatest mathematicians in history) by evaluating this series initially by a not-so-rigorous method. Later on, he gave alternative and more rigorous ways of getting the same result.

Prove that the series converges and gets an upper limit. Then, try to evaluate the series.


Due Nicolas Oresine:

Consider the following infinite series: \phi(s)=1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + ldots

We can re-write the preceding series as follows: \phi(s) = 1+ (\frac{1}{2^{s}}+\frac{1}{3^{s}}) + (\frac{1}{4^{s}} + \frac{1}{5^{s}} + \frac{1}{6^{s}} + \frac{1}{7^{s}}) + \ldots, which in turn is less than

1 + (\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots. Now, the RHS of this can be re-written as

1+(\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots=1 + \frac{1}{2^{(s-1)}}+ (\frac{1}{2^{(s-1)}})^{2} + \ldots, which is a geometric series and it is given by


Now, we can say that \phi(s) will converge if \frac{1}{2^{(s-1)}}<1 \Longrightarrow s >1.

In order to prove what is asked, we start with \phi(s)=1 + \frac{1}{2^{s}}+ \frac{1}{3^{s}}+ \frac{1}{4^{s}}+\ldots

And, then multiply both sides by \frac{1}{2^{s}} and then subtract the resulting equation from the preceding equation to get


where all the terms containing the reciprocals of the sth power of even numbers vanished.

Repeating this procedure with \frac{1}{3^{s}} gives

(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})\phi(s)=1+\frac{1}{5^{s}}+ \ldots

where all terms containing the reciprocals of the sth power of multiples of 3 vanished.

By continuing this with all prime numbers, we get

\prod_{p}(1-\frac{1}{p^{s}})\phi(s)=1, where p represents all prime numbers. Thus, we get

\phi(s)=1 + \frac{}{} + \frac{}{} + \frac{}{} + \ldots =\frac{1}{\prod_{p}(1-\frac{1}{p^{s}})}

This is a remarkable result because the LHS is concerned with only positive integers, whereas the RHS is concerned with only primes. This result is known as the “Golden Key of Euler”.

Riemann created his famous \zeta- function by extending the variable s to the entire complex plane, except s=1 withΒ 

\zeta(s)=1+ \frac{1}{2^{s}} + \frac{1}{3^{s}} + \ldots .

This function is now very famous as the Riemann zeta function.

How can we apply the Golden Key of Euler to Mengoli’s question that we started with?

Ans. In the Golden Key of Euler, substitute s=2.

Hence, we get the upper limit of the given series is 2.

Euler’s proof (1775):

The proof ran as follows:

It is a little roundabout way of arriving at the correct answer from a known result. Consider McLaurin’s series expansion of sin x:

\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} -\frac{x^{7}}{7!} + \frac{x^{9}}{9!} + \ldots

By dividing both sides by x and then substituting y=x^{2} on the right side, we get the following:

\frac{\sin{(x)}}{x} = 1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{7!} + \ldots

By taking a special value of x=n\pi (and, hence y=n^{2}\pi^{2}), we get the following:

\frac{\sin (n\pi)}{(n\pi)}=0=1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{5!}+ \ldots

NoteΒ  that preceding equation is not a polynomial, but an infinite series.Β But, Euler still treated it as a polynomial (that is why it was not accepted as a rigorous result) and observed that this “infinite” polynomial has roots equal to y_{n}=n^{2}x^{2}. Then, Euler had used the fact that the sum of the reciprocals of the roots is determined by the coefficient of the linear term (here, the y-term) when the constant is made unity. (check this as homework quiz, for a quadratic to be convinced).Β So, Euler had arrived at the following result:

1-\sum_{i=1}^{\infty}\frac{6}{y_{n}}=0 \Longrightarrow \sum_{i=1}^{\infty}\frac{1}{y_{n}}=\frac{1}{6}. With y_{n}=n^{2}(\pi)^{2}, we get the following:

\sum_{i=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}}=\frac{1}{6} or, \sum_{1}^{n^{2}}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}.

Another proof also attributed to Euler that uses the series expansion of sin (x) goes as follows below:

\sin {(x)} has roots given by 0, \pm \pi, \pm 2\pi, \pm 3\pi, …So does this polynomial that Euler reportedly constructed:


So, Euler considered the preceding equation to be equivalent to:

\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots=0

Then, he had equated the coefficient of x^{3} in both to get the result:

\sum_{n=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}} = \frac{1}{3!} = \frac{1}{6}.

Thus, \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}.

Later on, Euler had provided a few more alternate and rigorous proofs of this result.

Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.

Amazon India link:

Hope you all enjoyed it — learning to think like Euler !! By the way, it did take a long time for even analysis to become so rigorous as it is now….You might like this observation a lot. πŸ™‚ πŸ™‚ πŸ™‚

Nalin Pithwa.

Is there a “Fermat” in you? ? :-)

Fermat’s problem:

Pierre de Fermat (1601-65), the great French mathematical genius, is described as the “Prince of Amateurs” as he was not a professional mathematician and never published any work during his lifetime. As a civil servant, he served in the judiciary and spent his leisure with mathematics as his hobby. He corresponded with the best of mathematicians of his time. He teased the contemporary English mathematicians, Wallis and Digby, with the following problem, who had to admit defeat.


26 is a number that is sandwiched between a perfect square (25) and a perfect cube (27). Prove that there is no other such number.

Please do try and let me know your solutions. Even partial solutions are welcome.


Nalin Pithwa.

Inequalities and mathematical induction: RMO sample problems-solutions


1. Prove the inequality —- 2^{n}(n!)^{2} \leq (2n)! for all natural numbers greater than or equal to 1.

Proof 1:

First consider the following: 2.6. 10.14 \ldots (4n-2)=\frac{(2n)!}{n!}. Let us prove this claim first and then use it to prove what is asked: Towards, that end, consider

RHS = \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)\ldots 4.2.1}{\ldots (n-1)n}=LHS, cancelling off the common factors in numerator and denominator of RHS. (note this can also be proved by mathematical induction! πŸ™‚ )

In the given inequality:

we need to prove 2^{n}(n!)^{2} \leq (2n)!

consider \ldots (4n-2)=\frac{(2n)!}{n!} where

LHS = (2.1)(2.3) (2.5) (2.7) \ldots 2(n-1), this is an AP with first term 2 and nth term (4n-2) and common difference 4; there are n factors “2”; hence, we 2^{n} \ldots (n-1)=\frac{(2n)!}{n!} so we get

2^{n} (n!)\ldots (n-1) = (2n)!; multiplying and dividing RHS of this by\ldots n, we get the desired inequality. Remember the inequality is less than or equal to.

Problem 2:

Establish the Bernoulli inequality: If (1+a) > 0, then (1+a)^{n} \geq 1+na.

Solution 2:

Apply the binomial theorem, which in turn, is proved by mathematical induction ! πŸ™‚

Problem 3:

For all natural numbers greater than or equal to 1, prove the following by mathematical induction:

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{n^{2}} \leq 2-\frac{1}{n}

Proof 3:

Let the given proposition be P(n).

Step 1: Check if P(1) is true. Towards that end:

LHS=\frac{1}{1^{2}}=1 and $latex RHS=2-\frac{1}{1}=2-1=1$ and hence, P(1) is true.

Step 2: Let P(n) be true for some n=k, k \in N. That is, the following is true:

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} \leq 2 -\frac{1}{k}

Add \frac{1}{(k+1)^{2}} to both sides of above inequality, we get the following:

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} + \frac{1}{(k+1)^{2}} \leq 2-\frac{1}{k}+\frac{1}{(k+1)^{2}}

Now, the RHS in above is 2-\frac{1}{k} +\frac{1}{(k+1)^{2}}=2-\frac{k^{2}+k+1}{k(k+1)^{2}}. We want this to be less than or equal to 2-\frac{1}{k+1}. Now, k \in N, k>1, so what we have to prove is the following:

-\frac{k^{2}+k+1}{k(k+1)^{2}} \leq -\frac{1}{k+1}, that is, we want to prove that

(k+1)(k^{2}+k+1) \geq k(k^{2}+2k+1), that is, we want k^{3}+k^{2}+k+k^{2}+k+1 \geq k^{3}+2k^{2}+k, that is, we want k+1 \geq 0, which is obviously true. QED.


Nalin Pithwa.