IIT JEE , AIEEE Forum - Mathematics , Algebra - progressions,,,,,please urgentt .............

shobhita

.prove that (xyz)^(1/3)[1/x + 1/y + 1/z ]>=3,where x,y,z belongs to R

swaroopcool21

hey, this is simple checkout.........

x, y, z are any real nos (as given)

their GM = (x+y+z) ^ (1/3)

HM = 3 / [ 1/x + 1/y + 1/z ]

we have

GM >= HM

so,

(x+y+z) ^ (1/3) >= 3 / [ 1/x + 1/y + 1/z ]

transporting [1/x + 1/y + 1/z]

on other side,

we get,

(x+y+z)^1/3 { 3 / [ 1/x + 1/y + 1/z ]} >=3

waiting for salute mann

kusum

hi ,

applying am >= gm for 1/x,1/y,1/z
we get,
[1/x + 1/y +1/z]/3 >=(xyz)^(-1/3)

so corss multiply ,u get ur result .

JimboJones

The inequality is not valid for all x,y,z

R

Take x = -1/3, y = -1/2 and z = 1/6

1/x + 1/y + 1/z = 1

(xyz)^(1/3) = (1/36)^(1/3)

Thus (xyz)^(1/3)[1/x + 1/y + 1/z ]>=3

=> (1/36)^(1/3) >=3 which is not true.

For the inequality to be valid x,y,z

R+

The solutions provided for this question does make this assumption (otherwise you need to change the sign of the inequality when you multiply -ve numbers).

swaroopcool21

ya

jimbo jones

is right the ques. must specify that the 3 nos. are posive real nos.

else the ques. is cracked........................

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