IIT JEE , AIEEE Forum - Mathematics , Algebra

sachin_gupta1991

Prove that if x is real, the minimum value of (a+x)(b+x)

---------------

(c+x)

for x>-c , a>c, b>c

is

(

a-c +

b-c)².

shine

see.........we can solve dis using derivatives
dy/dx=( (x+c)(2x+a+b) -(x^2 +ax+bx +ab))/(x+c)^2

but for min or max dy/dx=0
=> on simplifying
x^2 +2cx+ac +bc -ab=0
=>x^2+2cx+c^2 -c^2 +ac +bc -ab=0
=>(x+c)^2= c^2 -ac -bc +ab
=> x= +-((a-c )(b-c))^1/2 -c

since x>-c
thus x= ((a-c )(b-c))^1/2 -c

now at dis pt d fnc attains min value

so d value is

(a-c + (a-c )(b-c))^1/2) * (b-c +(a-c )(b-c))^1/2)/ (a-c )(b-c))^1/2
=(a-c)^1/2 ((a-c)^1/2 + (b-c)^1/2) * (b-c)^1/2 ((a-c)^1/2 + (b-c)^1/2)
__________________
(a-c )(b-c))^1/2

= ((a-c)^1/2 + (b-c)^1/2)^2

hope ths correct soln

plzz rate if it helps u....................

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