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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 \( a+b+c=ab+bc+ca \)
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]
 one year ago
 one year ago
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 \( a+b+c=ab+bc+ca \) \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]
 one year ago
 one year ago

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mukushlaBest ResponseYou've already chosen the best response.2
man u most be on tensor and pde now...lol :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
on ... just for fun. if possible quick ... man
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
just this is bugging me so badly Exercise 1.1.8.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Ctrl + F ... search 1.1.8. Don't have much time to waste
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Actually, this question is very easy.. @experimentX
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Did not you get the question clearly ?? It is saying to prove that thing..
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
how do you prove that?
 one year ago

waterineyesBest 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..
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
already wasted 1 hrs ... can't waste another ... lol
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Actually, that pdf is quite useful for me.. Thanks for giving me that..
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
some fool upvoted me :D ... that doesn't work at all .... lol
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
yeah ... i know that ... just need to prove it.
 one year ago

experimentXBest 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.
 one year ago

experimentXBest 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}]
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
crazy crafty manipulation
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
the second one ... i can prove half of the original problem without it.
 one year ago

experimentXBest 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
 one year ago

experimentXBest 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}]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
what kind of problems is this x + 3n/x ... for x >= 3
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
can't ...have to do without it.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
don't know what property is used here ..
 one year ago

mukushlaBest 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\[\frac{ab^{3}}{c}+\frac{bc^{3}}{a}\ge 2b^{2}c\]...
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
which book do you use man??
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
oh sorry man original book is in persian and not translated to other languages :(
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
this is 500 pages about geometry and inequalities
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
you follow a very good writer!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
see ya :) thanks for help man
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Are you Linkha Santosh @experimentX ?? Sorry if there is any spelling mistake in name..
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
no mistake ... the first is my surname.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Santosh Linkha ..??
 one year ago

waterineyesBest 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..
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
lol ... what words?
 one year ago

waterineyesBest 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..
 one year ago

experimentXBest 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"
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Ha ha ha... Yeah, this is true one..
 one year ago
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