106a+1/b+b+1/c+c+1/a = (a+b+c)+(1/a+1/b+1/c) ≥ 6 (AM-GM)
[ a+b+c ≥ 33√abc=3 and 1/a+1/b+1/c ≥ 33√1/abc = 3]
(AM-GM) again
3√(a+(1/b)-1)(b+(1/c)-1)(c+(1/a)-1) ≤ (a+1/b+b+1/c+c+1/a-3)/3
So, 3√(a+(1/b)-1)(b+(1/c)-1)(c+(1/a)-1) ≤ (6-3)/3 = 1
a,b,c are positive reals with abc=1
Prove that
(a+(1/b)-1)(b+(1/c)-1)(c+(1/a)-1)≤1
-
UP 0 DOWN 0 0 15
15 Answers
106
1you said a+b+c+1/a+1/b+1c=/c >=6
you wrote
3√(a+(1/b)-1)(b+(1/c)-1)(c+(1/a)-1) ≤ (a+1/b+b+1/c+c+1/a-3)/3
but then (a+1/b+b+1/c+c+1/a-3)/3 >=1 not <=1 so the expression to the left need not be always less than equal to 1 It can be greater than one and less than the right exp. too ..
106rohan (a+1/b+b+1/c+c+1/a-3)/3 ≥ LHS of waht i wrote
as a+1/b+b+1/c+c+1/a ≥ 6 So, a+1/b+b+1/c+c+1/a-3 ≥3
So, if i put in first one:
≥3/3 ≥ LHS
So, LHS ≤ 1 isnt it?
1ok ashish you wrote (a+1/b+b+1/c+c+1/a-3)/3 >=1 so let it be 1.5
then you get
3√(a+(1/b)-1)(b+(1/c)-1)(c+(1/a)-1) <=1.5
so it can be 1.2 also which is greater than 1 !
9another IMO easy one ??????
but rohan better post these in olympiad
as far as jee is concerned they r not of much use
1This doesnt require any extra knowledge of inequalities as far my proof is concerned ..
1just substitute a=(x/y) , b=(y/z) , c=(z/x)
to prove,
(x+y-z)(x+z-y)(y+z-x) ≤ xyz
but this is true if xyz are sides of triangle, !!!
is their ny other condition on a,b,c ?
9is it so easy ???
im unable to do using AM-GM :( (its the only algebraic inequality in jee)
9dimensions uve already proved it !!
if x,y,z are sides of triangle ie(x+y-z),(x+z-y),(y+z-x)>0 then true
else if one of these is ≤0 then also true as LHs≤0 while RHS >0
( other cases like 2 of these are - ve is impossible as these would lead to
one of the sides to be ≤0
eg x+z ≤ y
y+z ≤ x
imply z≤0 which is false
1Yes celestine you completed his proof :)
take x+y-z=a
y+z-x=b
z+x-y=c
x+y+z= a+b+c
so , z=(b+c)/2
x=(a+c)/2
y= (a+b)/2
so we have to prove
abc≤(a+b)(b+c)(c+a)/8
but we have (a+b)/2 ≥√ab
similarly the others ≥√bc and √ca
so multiplying them we have the thing ≥abc Proved ..
So good work dimensions and celestine
btw my method was different ;
there can be two possibilities
a+b+c ≥1/a+1/b+1/c (where a b c same as in original equation )
or
a+b+c ≤1/a+1/b+1/c
the original expression is -
(a+1/b-1)(b+1/c-1)(c+1/a-1)
now if one term is negative then the expression becomes negative and hence less than 1
but when two terms are negative
we have eg. a+1/b-1<0 and b+1/c-1 <0
so we have a<1 and b<1 and 1/b<1 => b>1 contradiction;
so all terms need to be positive
Now if a+b+c ≥ 1/a+1/b+1/c
multiply each term by respectively b,c,a (and further abc=1)
= (ab+1-b)(bc+1-c)(ac+1-a)
=(1/c+1-b)(1/a+1-c)(1/b+1-a) = s
applying A.M ≥G.M
3√s≤(1/a+1/b+1/c-a-b-c+3)/3
but the (1/a+1/b+1/c-a-b-c+3)/3 <=1 as
we get
1/a+1/b+1/c ≤a+b+c which is true as per assumption
now if
a+b+c≤1/a+1/b+1/c then
multiply the terms in original expression respectively with 1/a,1/b and 1/c
we get
(1+1/ab -1/a)(1+1/bc-1/b)(1+1/ac-1/c)
= (1+c-1/a)(1+a-1/b)(1+b-1/c) = s
now again apply A.M≥G.M
3√s≤(a+b+c-1/a-1/b-1/c+3)/3
But (a+b+c-1/a-1/b-1/c+3)/3 <=1 as we get a+b+c ≤ 1/a+1/b+1/c which is again true as per assumption
!
so proved both ways ;