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Grade: 12th pass
        
Sir question number 4.......................................................
2 months ago

Answers : (1)

Aditya Gupta
1690 Points
							
as the line passes thru (a, b) we have
ad+bc= cd= k (say)
now, we apply AM-GM on ad and bc
(ad+bc)/2 is greater than or equal to sqrt(abcd)
k/2 is greater than or equal to sqrt(abk)
sqrt(k) is greater than or equal to 2*sqrt(ab)
now square both sides
k is greater than or equal to 4ab
or cd is greater than or equal to 4ab.
So minimum value of cd is 4ab and occurs when c= 2a and d= 2b.
now, we see that d= bc/(c – a) and as d is greater than 0, so c – a is greater than 0.
now, c+d= c + bc/(c – a)= c + b + ab/(c – a)= (a+b) + (c – a) + ab/(c – a)........(1)
now, apply AM-GM on (c – a) and ab/(c – a), we have
[(c – a) + ab/(c – a)]/2 is greater than or equal to sqrt(ab)
or (c – a) + ab/(c – a) is greater than or equal to 2*sqrt(ab)
so from (1) c+d is greater than or equal to (a+b) + 2*sqrt(ab)= (sqrt(a) + sqrt(b))^2
hence minimum value of c+d is (sqrt(a) + sqrt(b))^2 which occurs when (c – a)= ab/(c – a) or c= a+sqrt(ab) and the corresponding value of d.
kindly approve :=)
2 months ago
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