Cauchy Schwarz Inequality  Hemant Malhotra

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Cauchy Schwarz Inequality :
If A_{1},A_{2},A_{3}......A_{n} are n positive real number and B_{1},B_{2},B_{3}......B_{n} are n positive real numbers then
(A_{1}B_{1 }+ A_{2}B_{2 + ... + }A_{n}B_{n})^{2 }≤ (A_{1}^{2}×A_{2}^{2}×A_{3}^{2}......A_{n}^{2} ) x ( B_{1}^{2} × B_{2}^{2 }× B_{3}^{2} ... x B_{n}^{2})
If a^{2} + b^{2} + c^{2} = p^{2}+ q^{2} + r^{2} = 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 theseApply CauchySchwarz Inequality
(a^{2} + b^{2} + c^{2} ) (p^{2}+ q^{2} + r^{2}) > = (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 (x^{2}+y^{2})^{1/2}
A) 17
B ) 14
C) 16
D) 15Method1 Apply Cauchy
(15 * x + 20 * y)^{2 } < = (15^{2 }+ 20^{2}) (x^{2 }+ y^{2})
375^2 < = (625) (x^{2} + y^{2})
(x^{2} + y^{2}) > = (375/25)^2
so (x^{2} + y^{2})^{1/2} > = 15
Method 2 Alternate 15x + 20y = 375 so 3x+4y = 75
we need min value of (x^{2}+y^{2})^{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/(3^{2}+4^{2})^{1/2 } = 15
If x^{2 }+ y^{2} =14x + 6y + 6, what is the largest and minimum possible value that 3x + 4y can have
Method 1 Cauchy theorem
x^{2 }+ y^{2}  14x 6y  6 = 0
(x7)^{2} + (y3)^{2} = 64
Now apply Cauchy
we want 3x + 4y
and we have value of (x7)^{2} and (y3)^{2}
so break 3x+4y in x7 and y3 form to apply Cauchy
( 3 * (x 7 ) + 4 * (y3))^2 < = ( 3^{2} + 4^{2}) * ( (x  7)^{2} + (y  3)^{2} )
(3x+4y33)^{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))
x^{2} + y^{2 } 14x  6y  6 = 0(x7)^{2} + (y3)^{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(a^{2}+b^{2})
=  (3 * 7 + 4 * 3  k) / root(3^{2}+4^{2}) = 8
so 33k = 40
so 33k = + 40
so k = 7 or 73
Method 3 
x^{2} + y^{2 } 14x  6y  6 = 0 , let 3x+4y=k so x= (k4y)/3
put this value in equation  > (k4y/3)^{2}+y^{2}14((k4y)/36y6=0
now form quadratic in y and then D > =0 and u will find range of k
Method 4  x^{2} + y^{2 } 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 (24^{2 }+ 32^{2})
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 a^{2}+b^{2}+c^{2}+d^{2}+e^{2} = 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 < = (b^{2}+c^{2}+d^{2}+e^{2})*( 1^{2}+1^{2}+1^{2}+1^{2})
(8a)^{2} < = (16a^{2})*4
so 5a^{2}16a < =0
a (5a16) < =0
0 < =a < =16/5