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experimentX
 3 years 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 \( a+b+c=ab+bc+ca \)
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]
experimentX
 3 years 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 \( a+b+c=ab+bc+ca \) \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0man u most be on tensor and pde now...lol :)

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, this question is very easy.. @experimentX

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did not you get the question clearly ?? It is saying to prove that thing..

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0how do you prove that?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, that pdf is quite useful for me.. Thanks for giving me that..

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0man second one's RHS will simplify to\[a^2+b^2+c^2\]

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0what 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}]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for the first one use AM_GM 4 times\[\frac{a^2}{b}+b\ge 2a\]\[\frac{b^2}{c}+c\ge 2b\]...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0crazy crafty manipulation

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry man i was out...how u realte this to second one?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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\[\frac{ab^{3}}{c}+\frac{bc^{3}}{a}\ge 2b^{2}c\]...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry man original book is in persian and not translated to other languages :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is 500 pages about geometry and inequalities

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

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0no mistake ... the first is my surname.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That 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..

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0lol ... what words?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ha ha ha... Yeah, this is true one..
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