INEQUALITIES

lets see who can do this. this one`s for all oldies in this site like priyam ,skygirl.eureka,srinath, prophet sir, celestine, anirudh, sankara,unique,mak, NISHANT BHAIYAN, MANISH BHAIYAN, LOKESH BHAIYAN and others. { i miss rohan and ith power}

let a,b,c≥0 satisfying ab+bc+ca=1

prove that Σ 1/√a2+bc ≥2√2

there the square root is for total denominator. ALL THE BEST.

31 Answers

24
eureka123 ·

only IMO questions are not dificult dear...........there are putnam ,berkeley,chinese olympiad nd russian olympiads which also have far difficult problems.....[1]

i am trying this..........will try to get in 4 days[9]

11
Subash ·

ab+bc+ca=abc

is this given or assumed by you bhaiyya

1
skygirl ·

@mathie.... duniya badal jaye... no compromising wid sleep... [2]

:P sry [3]

62
Lokesh Verma ·

A couple of changes to make the proof more elaborate... Only one line added and one line (which i wrote wrongly removed)

apply AM GM inequality.

equality holds when
a2 + bc = b2 + ac
thus, (a-b)(a+b)=(a-b)c so a=b or a+b=c
similarly, (a-c)(a+c)=(a-c)b so a=c or a+c=b
and (c-b)(c+b)=(c-b)a so c=b or b+c=a

assume a=b

then the three cases will reduce to a=b
and ( a=c or c=0 )

if a=b and a=c

then we have proved the result.

if a=b and c=0

then ab=1 so a=1 and b=1 and c=0 which holds bcos LHS =3 <2 root 2

now take the second case

a+b=c and a+c=b and b+c=a

cos if any other holds (i mean a=c or b=c) we will be back to our previous case.

so a+b+c=0 which is not true. hence not possible.

Is it done? Or you think there are flaws?

1
palani ............... ·

would rationalising each term be useful???????????

39
Dr.House ·

tell me something, i accept it will take half hour but cant u take out half our from your sleep and do this?

its not neccesity but this question is worth it. nothing other than simple inequalities, yet a great question. thats why i siad lets see who can do this.

1
skygirl ·

fully agree wid celestine...

39
Dr.House ·

what? so no one able to do this or just left it off.?????????

1
sindhu br ·

If the root was not for whole denominator .............it would be better....
[7]

9
Celestine preetham ·

a prob like this wud eat away at least 1/2 hr so u shud try this in 11th itself

9
Celestine preetham ·

this isnt a jee prob so cant compare !!

probs like these are very interesting but time consuming
i ud advice to concentrate on jee maths for next 2 months cos it requires speed

11
Anirudh Narayanan ·

Sorry dude....this problem is out of my scope.....can anyone tell me how this problem compares to JEE level of toughness?

39
Dr.House ·

ANIRUDH I THINK IT WOULD BE BETTER IF U CONCNTRATE ON GETTING THE SOLUTION

11
Anirudh Narayanan ·

hehe......I am oldie [3][3][3]

9
Celestine preetham ·

heres a very stupid sol ;)

the a=b=c gives maxima ie 3√3/2

now minima by anti symmetry rule occurs when a tends to maxima with b,c tending to minima ie b,c→0 and a→1/2(0) sub and solve

39
Dr.House ·

its not about getting in 4 days , its about striking the excellent way to do it.

1
Philip Calvert ·

[12]

39
Dr.House ·

no sir, AS FAR AS I KNOW ITS NOT AN IMO QUESTION, IF NOT EUREKA WOULD HAVE ALREADY POINTED na .

really as far as i know its not an imo mone.

but i really loved the way i had to get to its solution. even i had to try for 4 days.

341
Hari Shankar ·

didnt say i know how to. could u give some more time

39
Dr.House ·

u r right prophet sir. by the way no oldies who can do this?{excluding prophet sir, he is a math genious }

341
Hari Shankar ·

i does. just let a,b→0

1
? ·

please check the question once more bhargav ...does the inequality holds ?

1
? ·

hehe ......sleepy brain !

33
Abhishek Priyam ·

there the square root is for total denominator. ALL THE BEST.

he wrote that...

1
? ·

bhargav is the whole denominator under root ?

39
Dr.House ·

damn sure

9
Celestine preetham ·

i think q is mistyped are u sure its ≥ ?

9
Celestine preetham ·

by law of symmetry put a=b=c=1/√3 and solve

39
Dr.House ·

areh symmetrically change a, b, c

33
Abhishek Priyam ·

Σ 1/√a2+bc ≥2√2

kiska sum hai... which is variable for Σ [12]

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