IIT JEE , AIEEE Forum - Mathematics , Algebra

anchitsaini

if x,y,z>0

prove that

2/(x+y) + 2/(y+z) + 2/(z+x) >= 9/(x+y+z)

akhil_o

let x+y=a
x+z=b
y+z=c

using AM>=HM

3abc/(ab+bc+ca)<=a+b+c/3

rearranging

9/a+b+c<=ab+bc+ca/abc

9/(a+b+c)<=1/a+1/b+1/c

a+b+c= 2(x+y+z)

1/2(2/a+2/b+2/c)>=9/2(x+y+z)

or

2/a+2/b+2/c>=9/x+y+z

2/(x+y)+2/(y+z)+2/(x+y)>= 9/(x+y+z)

hence proved

anchitsaini

well done but there is another method that i am looking for
and it is really short

Greatdreams

1/x+y + 1/y+z + 1/z+x is in L.H.S

or we have (x+y)^-1 + (y+z)^-1 + (z+x)^-1

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

So R.H.S = 3 [2(x+y+z)/3] ^-1

or 3 [ 3/ 2(x+y+z) ] = 9/2(x+y+z)

So 2 [ 1/x+y + 1/y+z + 1/z+x ]

9/(x+y+z)

Inequality used :

(x ^m + y ^m + z ^m + ...............) / n

[(x+y+z........)/n] ^m

anchitsaini

we are seeing good methods here
but there is one more way
lets see who does it by that method

Greatdreams

We can use anothr one

x+y+z = a

So a - x , a - y , a - z

0

So A.M

H.M

We have ,

[3a - (x+y+z)] / 3

3/ [ 1/a-x + 1/a-y + 1/a-z ]

or 2a/3

3/ [ 1/a-x+ 1/a-y+1/a-z)]

or 1/a-x+ 1/a-y+1/a-z

9/2a

or 1/x+y + 1/y+z + 1/z+x

9/2a

I am not totally sure whether we can use the

sign.......

I have a little doubt over that coloured step.

anchitsaini

still
there is another (here is the hint)
very short method

Greatdreams

Are you thinking of this way that since x,y,z > 0,

So we can consider that they are sides of a triangle and that

the sum of two sides is greater than the third side?

and then do the remaining?

or we can use this as well (x + y + z)(1/x + 1/y + 1/z)

9

I can't think of any shorter way!

anchitsaini

he he
no man

anchitsaini

should i post the method??

akhil_o

yeah plz do

nadeemoidu

consider the nos. 2/(x+y) , 2/(y+z) and 2/(z+x)

AM>=HM

(2/(x+y) + 2/(y+z) + 2/(z+x) ) / 3 >= 3/(x+y+z)

that is it . direct

saurabh_reincarnated

ya nadeem is rit...shortest 1

do we hve to only use A.M. > G.M.> H.M. only... in these type of prob...

isnt there any other methud

anchitsaini

all of u have done it the right way so pls don't shout at me. i had also done it through
A.M.>=H.M. but i found a new method to do this problem hence i am sharing it with u. (all the above methods are excellent and the method i am posting is nothing great)----
heres it--
using lemma which states that
x^2 + y^2 >=   (x+y)^2
---       ----         ---------
a         b            a+b

we can change the eqn to
      1      +        1     +       1
----------      ---------      ---------
(x+y)/2       (y+z)/2      (z+x)/2

which is
>=
(1+1+1)^2
--------------                                    which is the answer
[(x+y+y+z+z+x)/2]

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