2011/001) a problem in inequality

Let a,b,c,d be positive real numbers such that abcd = 1. Show that,?
(1 + a)(1 + b)(1 + c)(1 + d) >= 16

proof:

as a is real

(1+a) = (1-sqrt(a))^2 + 2 sqrt(a) or

1+ a >= 2 sqrt(a) ( we can also show it by AM GM inequality)

similarly

(1+b) > = 2 sqrt(b)

(1+c) > = 2 sqrt(c)

(1+d) > = 2 sqrt(d)

by multiplying

(1+a)(1+b)(1+c)(1+d)>= 16 sqrt(abcd) or > 16 as abcd = 1

2010/065) an example of chinese remainder theorem

What would be the least total no. which give the remainder 1,2 and 3 when divided by 7,9 and 11?

1,2 and 3 when divided by 7,9 and 11?

hence
x= 1 mod 7
x= 2 mod 9
x =3 mod 11

now 7 ,9 and 11 are pair wise coprime

x = 1 a1b1 + 2 a2b2 + 3 a3b3 mod 7*9*11(or 693)


where a1 = 9*11 = 99
a2 = 7 * 11 = 77
a3 = 7 * 9 = 63

and
a1b1 = 1 mod 7
a2b2 = 1 mod 9
a3b3 = 1 mod 11

a1 = 99 so b1 = 99 mod 7 or 1 mod 7 so b1 = 1

77b2 mod 9 = 1 so 5b2 mod 9 = 1 so b2= 2 (as 5*2 = 10 mod 9)

63b3 = 1 mod 11 or 8b3 =1 mod 11 so b3 = 7 as 7 * 8 = 56 = 1 mod 11

So x = 1 * 99 * 1 + 2 * 77 * 2 + 3 * 63 * 7 mod 693 = 1730 mod 693

= 344 mod 693

So x if of the form 693n + 344 or lowest x = 344

2010/064) if a+b+c=1, ab+bc+ca=2, abc=3, then

find the value of 1/(a+bc)+1/(b+ca)+1/(c+ab).


we have

a+b+c=1 ...1

ab+bc+ca=2 .. 2

abc=3 ...3

so a b c are roots of equation

x^3-x^2 + 2x - 3 = 0

further

as a+ b+ c = 1

so a = 1- (b+c)
so a + bc = 1-(b+c) + bc = (1-b)(1-c)

similarly b+ ca = (1-c)(1-a)
c + ba = (1-b)(1-a)

we need to find 1/(a+bc)+1/(b+ca)+1/(c+ab) = 1/ ((1-b)(1-c)) + 1/ ((1-c)(1-a)) + 1/ ((1-b)(1-a))

= (1-a) + (1-b) + (1-c)/(1-a)(1-b)(1-c) = ((a-1) + (b-1) + (c-1))/(a-1)(b-1)(c-1)



so we need to form an equation whose roots are a-1 , b- 1 and c- 1

as a b and c are roots of f(x) = x^3-x^2 + 2x - 3

so a-1 b- 1 and c- 1 are roots of

f(x+1) = (x+1)^3 - (x+1)^2 + 2(x+1) - 3 = x^3 + 2 x^2 + 3x -1

so (a-1) + (b-1) + (c-1) = - 2

(a-1)(b-1)(c-1) = 1

so given expression = ((a-1) + (b-1) + (c-1))/((a-1)(b-1)(c-1)) = - 2

2010/063) prove the inequality a+b/c+b+c/a+c+a/b>6 for a,b,c > 0 .?

this can be proved in at least 2 ways


method 1
we need to prove

(a+b)/c+(b+c)/a+(c+a)/b>6

or (a+b+c)/c + (a+b+c)/a + (a+b+ c)/b > 6

or (a+b+c) ( 1/a + 1/b+ 1/c) > 9

we know by AM GM inequality (a+b + c)/ 3 >= (abc)^(1/3)

again (1/a+ 1/b+ 1/c)/3 > = (1/abc)^(1/3)

multiplying we get (a+b + c)/ 3 * (1/a+ 1/b+ 1/c)/3 > = 1

or (a+b + c) * (1/a+ 1/b+ 1/c) >= 9

it is equal when a = b = c

proved


method 2
x+ 1/x >= 2 as (x+1/x) = (sqrt(x) -1/sqrt(x))^2 + 2

puttiing x = a/b . b/c and c/a in 3 times we get 3 eqautions

a/b + b/a >= 2
a/c+ c/a > = 2
b/c + c/b > = 2
adding above 3 and rearranging the terms we get the result.

2010/062) What are all the rational numbers x such that x + 1/x is an integer?

x + 1/x = (x^2 + 1)/x

This will be an integer if x^2 + 1 = kx, for some integer k. Equivalently,

x^2 - kx + 1 = 0

==> x = [ k +/- sqrt(k^2 - 4) ] / 2


for this to be an integer sqrt(k^2 - 4) has to be an integer say h( same parity as k that is both odd or both even)

k^2-4 = h^2

k^2 - h^2 = 4

so (k-h)(k+h) = 4 = 1 * 4 and 2* 2 and -1 * -4 and -2 * - 2

so k+ h = k-h = 2 or k-h = k+ h = - 2

k+ h = k-h = 2 => k = 2 giving x = 1

k+ h = k-h = 2 => k = -2 giving x = -1

so x = 1 or - 1

2010/061) prove the equality 2 + tan1° tan2° = cot1° tan2°

We know

Tan 2° = tan (1°+1°) = 2 tan 1°/(1- tan ^2 1°)
So tan 2° – tan 2° tan ^2 1° = 2 tan 1°

dividing by tan 1° we get

So tan 2° / tan 1° – tan 2° tan 1° = 2
Or
tan 2° cot 1° – tan 2° tan 1° = 2

So

tan 2° cot 1° = 2 + tan 2° tan 1°

2010/060) Proof of Bezout Identity

Bezout Identity states that

If a and b are integers (not both zero) then there exists integers u and v such that

Gcd(a,b) = au + bv

(Note: u and v are not unique

For example

au + bv = a(u-b) + b(v+a))

this can be used by backtracking the euclid’s equation to find Gcd(a,b) in terms of a b. this can be found in a number of books and is the standard process.

However this can be proved using pigeon hole principle also as below

We know that au is divisible by Gcd(a,b) so au mod b is divisible by Gcd(a,b)

Let gcd(a,b ) = k and
b / gcd(a,b) = t

Now taking an mod b( n from 1 to b/gcd(a,b)-1 that is t -1 ) there are

t-1 remainders

they are kn1, kn2, kn3, so on

there are t-1 remainders and all are divisible by k

no remainder can be zero

because na (n from 1 to t-1) cannot be a product of b.

no 2 remainder can be same if they are then difference is a multiple of b

so there has to a n such that one of the remainder is k

as all t- 1 remainders has to be different and values from 1 to n-1 so one remainder has to be 1

an = k mod b or an + bm = k

(u = n, v= m satisfy the condition)