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Concept 2 Inequalities I

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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 !

Written by Implex

August 27, 2008 at 7:41 am

7 Responses

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  1. Appreciate this initiative of yours. Let’s see how much do I learn from you!!

    Deepshikha

    August 27, 2008 at 4:37 pm

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

    (n!)^3 = (1.2.3………n-2.n-1.n)^3

    We will group (n,1),(n-1,2) …… so that we get sum as n+1

    Now 2 cases arise
    1) n is even-

    n!^3 = [(1.n)(2.n-1)(3.n-2)……]^3
    =[sqrt(1.n)sqrt(2.n-1)…….]^6<= [ (n+1)/2 * (n+1)/2*…..n/2 times]^6 = [(n+1)/2]^3n
    =[(n+1)/2]^2n [(n+1)/2)]^n

    NOw, [(n+1)/2)]^n(n!)^3

    Correct?

    Amit

    August 30, 2008 at 4:26 pm

  3. Reposting-

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

    (n!)^3 = (1.2.3………n-2.n-1.n)^3

    We will group (n,1),(n-1,2) …… so that we get sum as n+1

    Now 2 cases arise
    1) n is even-

    n!^3 = [(1.n)(2.n-1)(3.n-2)……]^3
    =[sqrt(1.n)sqrt(2.n-1)…….]^6<= [ (n+1)/2 * (n+1)/2*…..n/2 times]^6 = [(n+1)/2]^3n
    =[(n+1)/2]^2n [(n+1)/2)]^n

    NOw, [(n+1)/2)]^n< n^n

    Hence the inequality is proved

    Similarly it can be proved when n is odd.

    Correct?

    Is there anyway to edit the replies, outtimed?

    Amit

    August 30, 2008 at 4:33 pm

  4. Practice Problem 2.1

    AM of 1st n natural numbers n(n+1)/2n >= (n!)^1/n =>(n+1/2)^2n >=(n!)^2.
    Also we know that n^n>n! for all natural numbers. Multiplying this to above inequality we get:

    (n^n)[(n+1)/2]^(2n)>(n!)^3

    Vineet Singh

    September 1, 2008 at 7:08 pm

  5. Practice Problem 2.2
    Since sum of two sides of a triangle is always greater than the third. So AM>=GM can be applied.

    {(x+y-z)+(y+z-x)+(z+x-y)}/3 >= ((x+y-z)(y+z-x)(z+x-y)}^1/3
    =>(x+y+z)^3>= 27(x+y-z)(y+z-x)(z+x-y)

    Vineet Singh

    September 1, 2008 at 7:16 pm

  6. Practice Problem 2.3
    AM of 1st n natural numbers n(n+1)/2n >= (n!)^1/n=> (n+1)^n>=2.4.6….2n

    Please correct the question. It should have ‘>=’ sign

    Vineet Singh

    September 1, 2008 at 7:20 pm

  7. Can somebody plz provide the solutions to the practice problem 2.4 and 2.5.

    Rahul

    March 29, 2009 at 1:52 pm


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