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.ย Clearly, the presence ofย 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.ย So, this is a beautiful application of arithmetic mean-geometric mean inequality twice. ๐Ÿ™‚ ๐Ÿ™‚

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 !!!ย BINGO! ๐Ÿ™‚ ๐Ÿ™‚ ๐Ÿ™‚

Nalin Pithwa.

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 0ย  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.