Cauchy Schwarz Inequality - Hemant Malhotra


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Cauchy Schwarz Inequality :

    If A1,A2,A3......An are n positive real number and B1,B2,B3......Bn are n positive real numbers then

    (A1B1 + A2B2 + ... + AnBn)2 (A12×A22×A32......An2 ) x ( B12 × B22 × B32 ... x Bn2)

    If a2 + b2 + c2 = p2+ q2 + r2 = 101, where a, b, c, p, q and r are all distinct real numbers, then which of the following inequalities is true?

    1) ap + bq + cr < 99
    2) ap + bq + cr < 101
    3) ap + bq + cr < 202
    4) ap + bq + cr < 200
    5) None of these

    Apply Cauchy-Schwarz Inequality

    (a2 + b2 + c2 ) (p2+ q2 + r2) > = (ap + bq + cr)2

    so (101) × (101) > = (ap + bq + cr)2

    Now distinct

    (101) ×(101) > (ap+bq+cr)2  so (ap+bq+cr) < 101

    15x + 20y = 375. what is the minimum value of (x2+y2)1/2

    A) 17
    B ) 14
    C) 16
    D) 15

    Method1- Apply Cauchy

    (15 * x + 20 * y)2 < = (152 + 202) (x2 + y2)

    375^2 < = (625) (x2 + y2)

    (x2 + y2) > = (375/25)^2

    so  (x2 + y2)1/2 > = 15

    Method 2- Alternate  15x + 20y = 375 so 3x+4y = 75

    we need min value of   (x2+y2)1/2 which is nothing but distance of (x,y) from (0,0)

    so basically we want to find perpendicular length to 3x + 4y = 75 from (0,0)

    75/(32+42)1/2  = 15

    If x2 + y2 =14x + 6y + 6, what is the largest and minimum possible value that 3x + 4y can have

    Method 1- Cauchy theorem

     x2 + y2 - 14x -6y - 6 = 0

    (x-7)2 + (y-3)2 = 64

    Now apply Cauchy

    we want 3x + 4y

    and we have value of (x-7)2 and (y-3)2

    so break 3x+4y in x-7 and y-3 form to apply Cauchy

    ( 3 * (x -7 ) + 4 * (y-3))^2 < = ( 32 + 42) * ( (x - 7)2 + (y - 3)2 )

    (3x+4y-33)2 < = (25) * (64)

    -40 < = (3x - 4y - 33) < = 5 * 8

     -40 < = 3x - 4y -33 < = 40

    -7 < = 3x - 4y < = 73

    So minimum  value is -7 and maximum  is 73

    Method 2- 2nd approach to solve this kind of questions ( question involving equation of circle))  

    x2 + y2 - 14x - 6y - 6 = 0

    (x-7)2 + (y-3)2 -64 = 0

    center = (7,3) and radius =8

    now let 3x + 4y = k  now this line will touch circle then only it will be maximum or minimum

    perpendicular from center to line will be equal to radius

    The length of perpendicular from a point (x1, y1) to a line ax + by + c = 0 is |(ax1 + by1 + c)/ root(a2+b2)|

    = | (3 * 7 + 4 * 3 - k) / root(32+42)| = 8

    so |33-k| = 40

    so 33-k = +- 40

    so k = -7 or 73

    Method 3 -    

    x2 + y2 - 14x - 6y - 6 = 0 , let 3x+4y=k so x= (k-4y)/3

    put this value in equation -- > (k-4y/3)2+y2-14((k-4y)/3-6y-6=0

    now form quadratic in y and then D > =0 and u will find range of k

    Method 4 - x2 + y2 - 14x - 6y - 6 = 0

    circle with centre 7,3 and radius 8

    7 + 8 * cos x = X

    3 + 8 * sin x = Y

    3 (7 + 8 cos x) + 4 (3 + 8 sin x)

    33 + 24 cos x + 32 sin x

    33 +/- sqrt (242 + 322)

    33 +/- 40 so -7 minimum and 73 maximum

    Given that the real numbers a,b,c,d and e satisfy simultaneously the relations a+b+c+d+e = 8 and a2+b2+c2+d2+e2 = 16, determine the maximum and minimum value of a.

    Method- we want value of a so apply cauchy in other terms and form inequality in terms of a

    so (b*1+c*1+d*1+e*1)^2 < = (b2+c2+d2+e2)*( 12+12+12+12)

    (8-a)2 < = (16-a2)*4

    so 5a2-16a < =0

    a (5a-16) < =0

    0 < =a < =16/5


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