prove arithmetic-geometric mean inequality for n = 3
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pka
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Suppose that each of a & b is a nonnegative number and \(\displaystyle a > b.\)
Then \(\displaystyle a_1 = \frac{{a + b}}{2}\quad \& \quad b_1 = \sqrt {ab}.\)
Thus we see that: \(\displaystyle \begin{array}{l}
a_1 < \frac{{a + a}}{2} = a\quad \& \quad b_1 > \sqrt {b^2 } = b \\
\frac{{a + b}}{2} \ge \sqrt {ab} \quad \Rightarrow \quad a > a_1 \ge b_1 > b \\
\end{array}\)
Do it again!
For n=2 define \(\displaystyle a_2 = \frac{{a_1 + b_1 }}{2}\quad \& \quad b_2 = \sqrt {a_1 b_1 }.\)
Then show that \(\displaystyle a > a_1 > a_2 \ge b_2 > b_1 > b.\)
Now you do it for n=3.
Then \(\displaystyle a_1 = \frac{{a + b}}{2}\quad \& \quad b_1 = \sqrt {ab}.\)
Thus we see that: \(\displaystyle \begin{array}{l}
a_1 < \frac{{a + a}}{2} = a\quad \& \quad b_1 > \sqrt {b^2 } = b \\
\frac{{a + b}}{2} \ge \sqrt {ab} \quad \Rightarrow \quad a > a_1 \ge b_1 > b \\
\end{array}\)
Do it again!
For n=2 define \(\displaystyle a_2 = \frac{{a_1 + b_1 }}{2}\quad \& \quad b_2 = \sqrt {a_1 b_1 }.\)
Then show that \(\displaystyle a > a_1 > a_2 \ge b_2 > b_1 > b.\)
Now you do it for n=3.
i don't quite get how i can use it to prove that for an x,y,z real numbers,(x+y+z)^3>= 3xyz !!
could u please give me a litle bit stronger hint !
thanks !!
sincerly,
yassine !
ps: of course, u're very welcome to give me simply the answer instead of the hint!
thanks
- Joined
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What have you tried? How far have you gotten? Where are you stuck?yaszine said:
I certainly hope you're not waiting for somebody to "do" your assignment for you. Most (legitimate) tutors don't do that.yaszine said:
Please reply showing everything you have tried so far. Thank you.
Eliz.
i don't actually understand why you want me to use a2 and b2 !! and how can i use the case for n=2 that you showed to prove the assertion for 3 !!
am i gonna assume that a>b>c or what ?! what i mean by i don't get what you mean or how i could use it, is that it doesn't make so much sense for me!!
and yeah ! of course i worked on it way too much before i actually got to post it in the forum ! and i'm still working on it ! but as i spent too much time on it, i think i deserve to know how it can be solved as the most important is that i, by now have the problem graved into my memory and so would be it's solution!
ok thanks for your help!
sincerly,
yassine
am i gonna assume that a>b>c or what ?! what i mean by i don't get what you mean or how i could use it, is that it doesn't make so much sense for me!!
and yeah ! of course i worked on it way too much before i actually got to post it in the forum ! and i'm still working on it ! but as i spent too much time on it, i think i deserve to know how it can be solved as the most important is that i, by now have the problem graved into my memory and so would be it's solution!
ok thanks for your help!
sincerly,
yassine
proved it !
oh i think i've proved ! it was quite late the time when i got to it but still !
i arrived at an expression saying 7xyz(2a^(1/n) + b^(1/n)) + 6 xyz !!
and so 2a^(1/n) + b^(1/n) tends to 3 as n goes to infinity (by applying the mean average arithmetic mean n times) thus the whole thing become >= 27xyz meaning that x+y+z >= 3 root of(xyz) !!
i hope my proof makes sens to you !
thank you for your help !!
sincerly,
yassine
ps: i didn't post my computations before i proved it cuz i was afraid i was in a wrong track ! but i wasnt!! :lol:
oh i think i've proved ! it was quite late the time when i got to it but still !
i arrived at an expression saying 7xyz(2a^(1/n) + b^(1/n)) + 6 xyz !!
and so 2a^(1/n) + b^(1/n) tends to 3 as n goes to infinity (by applying the mean average arithmetic mean n times) thus the whole thing become >= 27xyz meaning that x+y+z >= 3 root of(xyz) !!
i hope my proof makes sens to you !
thank you for your help !!
sincerly,
yassine
ps: i didn't post my computations before i proved it cuz i was afraid i was in a wrong track ! but i wasnt!! :lol:
it would be really helpful though if you know any website where i can exercise some binomial distrubtions including the methode of generating functions ! i have a midterm next tuesday and i'm really worried about it !
thanks very much for ur help again !
sincerly,
yassine!
thanks very much for ur help again !
sincerly,
yassine!