A community for students.
Here's the question you clicked on:
 0 viewing
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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2man 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

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

waterineyes
 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?

waterineyes
 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

waterineyes
 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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2man 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}]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2for 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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2sorry 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 ..

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2i 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??

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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2this 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

waterineyes
 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.

waterineyes
 3 years ago
Best ResponseYou've already chosen the best response.0Santosh Linkha ..??

waterineyes
 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?

waterineyes
 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"

waterineyes
 3 years ago
Best ResponseYou've already chosen the best response.0Ha ha ha... Yeah, this is true one..
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.