x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers $x, x_{1}, x_{2}, \ldots, x_{n}$ such that

$|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|$
…let us say this is given equality A

Solution:

Use the following inequality: $|a-b| \leq |a| + |b|$ with equality iff $ab \leq 0$

So, we observe that : $|\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|$ ,

Hence, LHS of the given equality is greater than or equal to:

$2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)$

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

$|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|$

So, applying the above to RHS of A:

$|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|$ .

But, RHS is equal to LHS as given in A:

That is, $|\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|$

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

$|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)$

The above two inequalities are like the following: $x \leq y$ and $x \geq 2y$ ; so what is the conclusion? The first inequality means $x2y$ or $x=2y$ ; clearly it means the only valid solution is $x=2y$ .

Using the above brief result, we have here:

$|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)$

Hence, we get $|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0$ , which in turn means that (by applying the definition of absolute value):

$|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|$ , which implies that $x_{1}=x_{2}= \ldots x_{n}=1$ .

Substituting these values in the given logarithmic absolute value equation, we get:

$n \times |\log{x}|+ n \times |\log{x}|=0$ , that is $2n \times |\log{x}|=0$ , and as $n \neq 0$ , this implies that $|\log{x}|=0$ which in turn means $x=1$ also.

Some random problems in algebra (part b) for RMO and INMO training

Posted by Nalin Pithwa

1) Solve in real numbers the system of equations:

$y^{2}+u^{2}+v^{2}+w^{2}=4x-1$

$x^{2}+u^{2}+v^{2}+w^{2}=4y-1$

$x^{2}+y^{2}+v^{2}+w^{2}=4u-1$

$x^{2}+y^{2}+u^{2}+w^{2}=4v-1$

$x^{2}+y^{2}+u^{2}+v^{2}=4w-1$

Hints: do you see some quadratics ? Can we reduce the number of variables? …Try such thinking on your own…

2) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers such that $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0$ and $\max_{1 \leq i <j \leq 5} |a_{i}-a_{j}| \leq 1$ . Prove that $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2} \leq 10$ .

3) Let a, b, c be positive real numbers. Prove that

$\frac{1}{2a} + \frac{1}{2b} + \frac{1}{2d} \geq \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$

More later

Nalin Pithwa.

Some random assorted (part A) problems in algebra for RMO and INMO training

Posted by Nalin Pithwa

You might want to take a serious shot at each of these. In the first stage of attack, apportion 15 minutes of time for each problem. Do whatever you can, but write down your steps in minute detail. In the last 5 minutes, check why the method or approach does not work. You can even ask — or observe, for example, that if surds are there in an equation, the equation becomes inherently tough. So, as a child we are tempted to think — how to get rid of the surds ?…and so on, thinking in math requires patience and introversion…

So, here are the exercises for your math gym today:

1) Prove that if x, y, z are non-zero real numbers with $x+y+z=0$ , then

$\frac{x^{2}+y^{2}}{x+y} + \frac{y^{2}+z^{2}}{y+z} + \frac{z^{2}+x^{2}}{x+z} = \frac{x^{3}}{yz} + \frac{y^{3}}{zx} + \frac{z^{3}}{xy}$

2) Let a b, c, d be complex numbers with $a+b+c+d=0$ . Prove that

$a^{3}+b^{3}+c^{3}+d^{3}=3(abc+bcd+adb+acd)$

3) Let a, b, c, d be integers. Prove that $a+b+c+d$ divides

$2(a^{4}+b^{4}+c^{4}+d^{4})-(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd$

4) Solve in complex numbers the equation:

$(x+1)(x+2)(x+3)^{2}(x+4)(x+5)=360$

5) Solve in real numbers the equation:

$\sqrt{x} + \sqrt{y} + 2\sqrt{z-2} + \sqrt{u} + \sqrt{v} = x+y+z+u+v$

6) Find the real solutions to the equation:

$(x+y)^{2}=(x+1)(y-1)$

7) Solve the equation:

$\sqrt{x + \sqrt{4x + \sqrt{16x + \sqrt{\ldots + \sqrt{4^{n}x+3}}}}} - \sqrt{x}=1$

8) Prove that if x, y, z are real numbers such that $x^{3}+y^{3}+z^{3} \neq 0$ , then the ratio $\frac{2xyz - (x+y+z)}{x^{3}+y^{3}+z^{3}}$ equals $2/3$ if and only if $x+y+z=0$ .

9) Solve in real numbers the equation:

$\sqrt{x_{1}-1} = 2\sqrt{x_{2}-4}+ \ldots + n\sqrt{x_{n}-n^{2}}=\frac{1}{2}(x_{1}+x_{2}+ \ldots + x_{n})$

10) Find the real solutions to the system of equations:

$\frac{1}{x} + \frac{1}{y} = 9$

$(\frac{1}{\sqrt[3]{x}} + \frac{1}{\sqrt[3]{y}})(1+\frac{1}{\sqrt[3]{x}})(1+\frac{1}{\sqrt[3]{y}})=18$

More later,
Nalin Pithwa

PS: if you want hints, do let me know…but you need to let me know your approach/idea first…else it is spoon-feeding…

Solutions to two algebra problems for RMO practice

Posted by Nalin Pithwa

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:

$a+b+c+ab+bc+ca+abc=7$ . Clearly, the presence of $a+b+c$ 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: $\frac{ab+bc+ca}{3} \geq (ab.bc.ca)^{1/3}$ . 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, $x \leq 1$ . So, this is a beautiful application of arithmetic mean-geometric mean inequality twice. 🙂 🙂

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 !!! BINGO! 🙂 🙂 🙂

Nalin Pithwa.

Inequalities and mathematical induction: RMO sample problems-solutions

Posted by Nalin Pithwa

Problem:

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

$RHS = \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)\ldots 4.2.1}{1.2.3.4\ldots (n-1)n}=LHS$ , cancelling off the common factors in numerator and denominator of RHS. (note this can also be proved by mathematical induction! 🙂 )

In the given inequality:

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

consider $2.6.10.14. \ldots (4n-2)=\frac{(2n)!}{n!}$ 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 $2^{n}1.3.5.7 \ldots (n-1)=\frac{(2n)!}{n!}$ so we get

$2^{n} (n!) 1.3.5.7.\ldots (n-1) = (2n)!$ ; multiplying and dividing RHS of this by $2.4.6.8.\ldots n$ , 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 ! 🙂

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.

Cheers,

Nalin Pithwa.

An easy inequality from Nordic mathematical contests !?

Posted by Nalin Pithwa

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” ! :-( :-) !

Posted by Nalin Pithwa

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.