we can find the ratios very easily as (y+z))/(x+y+z) ,(x+z)/(x+y+z) , (x+y)/(x+y+z) respectively where x,y,z are opposite sides to A,B and C respectively after that showing 2(x+y)(y+z)(x+z)>(x+y+z)^3 is quite easy assume x+y=a, y+z=b, x+z=c so 16abc> (a+b+c)^3 is all we have to prove (a+b+c) >= 3 cube root(abc) (a+b+c)^3>= 27 abc this gives me a feel of something wrong happening here equality holds when all are equal,a=b=c i.e it should be (x+y+z)^3 >=27/8 (x+y)(y+z)(z+x) we can check this also, let the triangle be equilateral, each ratio is 2/3, resultant is 8/27, question wrong , hence proved
I might be wrong, but i strongly feel i m not , please point out the mistake in the othercase :D
You Thought it in the right direction bhaiya.. you are missing somewhere yaar.. actually the thing you proved is the first part of the problem and i posted the second part.. well you used the right concept.. :D
I will give you a hint : Do something with the inradius it might help.
Well ek badiya tarika proof ka hai.. ye dekh lo !! it might be a fluke bt pata nahi..
take the triangle as a equilateral one.. then the thing will be 1.. that is greater than 1/2 and if the triangle is not an equilateral then we need a lill' bit adjustments.. which are not smaller than zero.. thats why for all triangles the inequality is true..
May be but you said that it is for the check but i think why cant we prove things with this help :D
check the figure! is it right? http://www.imaginationcubed.com/loader.php?aDrawingID=ef229029b2cdd071f8fc1190569d7ead&from_email=akashrupela%40ymail.com&from_name=akash+rupela
wow wow wow....devansh...u ve got a wrong queston!!!!!! the RHS of inequality is 1/4 not 1/2 ...as akash said check it for equilateral triangle..its 8/27<1/2...
nice one!!!! will hv a solution by tomorrow!
ReplyDeletearen't you supposed to tell what D E anf F are?
ReplyDeleteHey This was Just easy to guess.. ends of angle bisectors of A,B and C respectively.. hope it is much clear now .. !!
ReplyDeletewe can find the ratios very easily as (y+z))/(x+y+z) ,(x+z)/(x+y+z) , (x+y)/(x+y+z) respectively where x,y,z are opposite sides to A,B and C respectively
ReplyDeleteafter that showing 2(x+y)(y+z)(x+z)>(x+y+z)^3 is quite easy
assume x+y=a, y+z=b, x+z=c
so 16abc> (a+b+c)^3 is all we have to prove
(a+b+c) >= 3 cube root(abc)
(a+b+c)^3>= 27 abc
this gives me a feel of something wrong happening here
equality holds when all are equal,a=b=c
i.e it should be (x+y+z)^3 >=27/8 (x+y)(y+z)(z+x)
we can check this also, let the triangle be equilateral, each ratio is 2/3, resultant is 8/27, question wrong , hence proved
I might be wrong, but i strongly feel i m not , please point out the mistake in the othercase :D
@Akash Bhaiya:
ReplyDeleteYou Thought it in the right direction bhaiya.. you are missing somewhere yaar.. actually the thing you proved is the first part of the problem and i posted the second part.. well you used the right concept.. :D
well check the question then
ReplyDeleteIA IB IC/ AD BE CF is surely less than equal to 8/27 if A,B,C are vertices, and I is incenter.
I will give you a hint : Do something with the inradius it might help.
ReplyDeleteWell ek badiya tarika proof ka hai.. ye dekh lo !! it might be a fluke bt pata nahi..
take the triangle as a equilateral one.. then the thing will be 1.. that is greater than 1/2 and if the triangle is not an equilateral then we need a lill' bit adjustments.. which are not smaller than zero.. thats why for all triangles the inequality is true..
May be but you said that it is for the check but i think why cant we prove things with this help :D
can u please tell me how will the thing be 1 for equilateral triangle?
ReplyDeletecheck the figure! is it right?
ReplyDeletehttp://www.imaginationcubed.com/loader.php?aDrawingID=ef229029b2cdd071f8fc1190569d7ead&from_email=akashrupela%40ymail.com&from_name=akash+rupela
Nah Yaar.. not 1.. galat ho gaya.. yaar.. arey yaar.. bt i was explaining something with an xample.. i wasnt having a pen or a paper.. sorry !!!
ReplyDeleteWell yaar.. this might help..
ReplyDeleteWe know the inequality
(a+b-c)(a+c-b)(b+c-a)>0 .... (1)
where a,b,c are sides of triangle.
Now We should introduce elementary symmetric functions
u= a+b+c, v=ab+bc+ca w= abc... (2)
putting (2) in (1)
we will get something like
-u^3+4uv-8w >0...(3)
on the other hand
1/4 < f(a,b,c)...(4)
gives
-u^3+4uv-4w >0....(5)
(3) is obviously correct. so the other two are also correct.
So set a = y+z, b=z+x, c= x+y
with r=x/x+y+z, s=y/x+y+z t=z/x+y+z
AI/AD=1/2 (1+r)similarly we will get the other ones..
so f(a,b,c)= 1/8 (1+r)(1+s)(1+t)
putting the inequality in (4) completes the proof..
Correct me if wrong.. may be wrong on the symmetric functions side..
@ Akash bhaiya.. aapne first part prove kiya tha iss question ka bilkul correctly..
ReplyDeleteand yea.. wo jo 1 aaya tha wo shayad galti se maine ratio galat le liya..
wo 1 aa hi nahi sakta because numerator is smaller.. sorry !!
wow wow wow....devansh...u ve got a wrong queston!!!!!! the RHS of inequality is 1/4 not 1/2
ReplyDelete...as akash said check it for equilateral triangle..its 8/27<1/2...
yehi tha question mein yaar.. wo toh maine ab padha.. :D lol !! ye kya ho gaya.. Sorry guys.
ReplyDeleteOk.. yaar finally we are done with the question .. thanks all for replying.. :D
ReplyDeletelol
ReplyDelete