Cauchy Schwarz Inequality - Hemant Malhotra

[hemant_malhotra]

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 + A 32......An2 ) x ( B12 + B22 + B32 ... + 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