1if a \ge b \ge c from tchebychev
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \le \frac{16}{3} so maximum value = 16/3
from AM-GM \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge 3 so min value =3
\left(a+b+c \right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=16, a,b,c are strictly positive reals......Maximise and minimise
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}...
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15 Answers
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66@fibonacci
can you state the values of a, b, c for equality in each case?
1Kaymant sir i'm unable to find the values of a,b,c but is there any specific technique to do so?
also could you please post your method of finding the values of a,b,c
11fibonacci - ur answer is obviously wrong, as equality in A.M-G.M is when a=b=c, which is not possible here!
66The maximum value is \dfrac{1}{2}(13+\sqrt{5}) and the minimum value is \dfrac{1}{2}(13-\sqrt{5}). These are attained for infinitely many values of a,b,c. E.g. Take b=\dfrac{1}{2}(3+\sqrt{5})a and c=\dfrac{1}{2}(3-\sqrt{5})a for maximum; exchange b and c for minimum.
341I am curious as to the source of this problem. Is it that postal coaching stuff?
11No I saw this problem a few days back at mathlinks, but then lost its source - couldn't find it after last Tuesday!
341Hmm, its quite a difficult problem. Its proposed by a Vietnamese guy, so its more difficult than usual. I read the solution in his book and its quite a solution!
BTW, the thread is here: http://www.mathlinks.ro/viewtopic.php?search_id=1109338498&t=308229
1kaymant sir could you please tell the reason for chosing such values!!
66Well as far as I am concerned, I used the methods of Lagrange multipliers. Any body want that?
@theprophet
You have read the solution, could you verify the answer.
11Anant sir, I want that! I mean give ur soln....I don't know Langrange multipliers......PLS!!!