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experimentX
Group Title
this is bugging me:
for any \( a, b, c, d > 0\) prove that
\[ {a^2 \over b} + {b^2 \over c} + {c^2 \over d} +{d^2 \over a} \geq a+ b+c+d \]
Also if \( a+b+c=ab+bc+ca \)
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]
 2 years ago
 2 years ago
experimentX Group Title
this is bugging me: for any \( a, b, c, d > 0\) prove that \[ {a^2 \over b} + {b^2 \over c} + {c^2 \over d} +{d^2 \over a} \geq a+ b+c+d \] Also if \( a+b+c=ab+bc+ca \) \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]
 2 years ago
 2 years ago

This Question is Closed

mukushla Group TitleBest ResponseYou've already chosen the best response.2
man u most be on tensor and pde now...lol :)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
on ... just for fun. if possible quick ... man
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
just this is bugging me so badly Exercise 1.1.8.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
Ctrl + F ... search 1.1.8. Don't have much time to waste
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Actually, this question is very easy.. @experimentX
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Did not you get the question clearly ?? It is saying to prove that thing..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
how do you prove that?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Actually, I am talking about question.. Its solution is easy or not I don't know.. Ha ha ha..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
already wasted 1 hrs ... can't waste another ... lol
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Actually, that pdf is quite useful for me.. Thanks for giving me that..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
some fool upvoted me :D ... that doesn't work at all .... lol
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
man second one's RHS will simplify to\[a^2+b^2+c^2\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yeah ... i know that ... just need to prove it.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
olypiad problems ... i find it extremely difficult even at my level man!! i need to dig a hole hide.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
what kind of technique is this http://www.wolframalpha.com/input/?i=Minimize[{x+%2B+2+n%2Fx%2C+n+%3C%3D+3+%26%26+x+%3E%3D+3}%2C+{x}]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
for the first one use AM_GM 4 times\[\frac{a^2}{b}+b\ge 2a\]\[\frac{b^2}{c}+c\ge 2b\]...
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
crazy crafty manipulation
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
the second one ... i can prove half of the original problem without it.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
the last inequality i can get bu the use of quadratic formula ... but can't find any use of ... inequality n<=3
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
particularly I mean this http://www.wolframalpha.com/input/?i=Minimize[{x+%2B+3+n%2Fx%2C++x+%3E%3D+3}%2C+{x}]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
what kind of problems is this x + 3n/x ... for x >= 3
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
sorry man i was out...how u realte this to second one?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
can't ...have to do without it.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
don't know what property is used here ..
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
i found it from one my books i will copy it here \[(a+b+c)(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}) \geq (a+b+c)(a^2+b^2+c^2)\]\[\frac{ab^3}{c}+\frac{bc^3}{a}+\frac{ca^3}{b} \geq b^2c+c^2a+a^2b\]use AM_GM 3 times\[\frac{ab^{3}}{c}+\frac{bc^{3}}{a}\ge 2b^{2}c\]...
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
which book do you use man??
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
oh sorry man original book is in persian and not translated to other languages :(
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
this is 500 pages about geometry and inequalities
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
you follow a very good writer!!
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
see u later santosh
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
see ya :) thanks for help man
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Are you Linkha Santosh @experimentX ?? Sorry if there is any spelling mistake in name..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
no mistake ... the first is my surname.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Santosh Linkha ..??
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
That day you said now a days, everyone is on Facebook and I was just wondering why experimentX who said those words is not on Facebook..?? Now, it is clear to me..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
lol ... what words?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
I have written those.. "Everyone is on Facebook".. I think these you said to Yahoo that day..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
oh yeah ... kinda remembered on thing "if you are searched in Google instead of fB then you are successful person"
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Ha ha ha... Yeah, this is true one..
 2 years ago
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