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} ... + 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 these

Apply Cauchy-Schwarz 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) 15

Method1- 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

(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 < = ( 3^{2} + 4^{2}) * ( (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))

x^{2} + y^{2 }- 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(a^{2}+b^{2})|

= | (3 * 7 + 4 * 3 - k) / root(3^{2}+4^{2})| = 8

so |33-k| = 40

so 33-k = +- 40

so k = -7 or 73

Method 3 -

x^{2} + y^{2 }- 14x - 6y - 6 = 0 , let 3x+4y=k so x= (k-4y)/3

put this value in equation -- > (k-4y/3)^{2}+y^{2}-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 - 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})

(8-a)^{2} < = (16-a^{2})*4

so 5a^{2}-16a < =0

a (5a-16) < =0

0 < =a < =16/5

]]>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} ... + 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 these

Apply Cauchy-Schwarz 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) 15

Method1- 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

(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 < = ( 3^{2} + 4^{2}) * ( (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))

x^{2} + y^{2 }- 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(a^{2}+b^{2})|

= | (3 * 7 + 4 * 3 - k) / root(3^{2}+4^{2})| = 8

so |33-k| = 40

so 33-k = +- 40

so k = -7 or 73

Method 3 -

x^{2} + y^{2 }- 14x - 6y - 6 = 0 , let 3x+4y=k so x= (k-4y)/3

put this value in equation -- > (k-4y/3)^{2}+y^{2}-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 - 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})

(8-a)^{2} < = (16-a^{2})*4

so 5a^{2}-16a < =0

a (5a-16) < =0

0 < =a < =16/5

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