Problem of the Day 9th Oct 2011

In an increasing Arithmetic Progression, the product of the 5th term and the 6th term is 300. When
the 9th term of this A.P. is divided by the 5th term, the quotient is 5 and the remainder is 4. What is
the first term of the A.P.?
(a) 12                 (b) –40               (c) –16                     (d) –5

Concept 2 Inequalities I

Concept 2 Inequalities

Lets move on to our next concept, i.e Inequalities. Inequalities are generally present in cat and similar MBA papers, the question can be direct or indirect.

Concept 2.1 AM-GM Inequality

It means that AM( arithemetic mean) of  a set of positive numbers is always greater than or equal to the GM( geometric mean). The equality holds when the numbers are equal

(a+b+c)/3 >=(a+b+c)^(1/3)..........( 2.1)

Example 2.1 If a,b,c are positive numbers prove that (a+b)(b+c)(c+a)>=8abc

what we will do is  use AM-GM multiple times

(a+b)/2 >=sqrt(ab)

=>(a+b)>=2sqrt(ab)

similarly for others

(b+c)>=2sqrt(bc)

(c+a)>=2sqrt(ac)

then multiplying these three inequalities we get the desired result!

Practice Problem 2.1show that (n^n)[(n+1)/2]^(2n)>(n!)^3

Practice Problem 2.2 if x,y,z be the lengths of the sides of a triangle then prove that (x+y+z)^3>=27(x+y-z)(y+z-x)(z+x-y)

Practice Problem 2.3 show that for any natural number n, (n+1)^n>2.4.6....2n

Example 2.2 Show that for any natural number n 2^n>=1 +n.2^[(n-1)/2]

Lets see how we do this

2^n>=1+n.2^[(n-1)/2]

2^n-1>=n.2^[(n-1)/2] ( can you recognise the form?)

its the sum of a GP

we need to use AM-GM on the sum of GP

[1+2+2^2...+2^(n-1)]/n>(1.2.2^2...2^(n-1))^(1/n)

(2^n-1)/n> ( 2^(1+2+3..+n-1))^(1/n)=(2^[n(n-1)/2])^(1/n)=2^((n-1)/2)

so

2^n-1>2^((n-1)/2)

so we are done !!

Concept 2.2 Cauchy- Schwartz Inequality

If a,b,c and x,y,z be real numbers ( positive, negative or zero) then

(ax+by+cz)^2<=(a^2+b^2+c^2)(x^2+y^2+z^2)

Equality holds iff  a:b:c::x:y:z

Example 2.3 if x^4+y^4+z^4 =27 find min value of x^6+y^6+z^6

use cauchy on x^3,y^3,z^3 and  x,y,z

then (x^6+y^6+z^6)(x^2+y^2+z^2)>=(x^4+y^4+z^4)^2....(1)

use cauchy on the numbers x^2,y^2,z^2 and 1,1,1

then (x^4+y^4+z^4)(1+1+1)>=(x^2+y^2+z^2)^2

3(x^4+y^4+z^4)>=(x^2+y^2+z^2)^2...(2)

squaring both sides of 1 and using 2 we get

(x^4+y^4+z^4)^4<=3[(x^6+y^6+z^6)^2](x^4+y^4+z^4)

putting x^4+y^4+z^4=27 and taking positive square root we get

x^6+y^6+z^6>=81

Practice Problem2.4  if a,b,c be positive numbers such that a+b+c=4 find minimum value of a^3+b^3+c^3

Practice Problem 2.5 Find the min value of 2x+y if xy=8 and x,y are positive numbers

For any queries, post your doubts here itself !