Question

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

Answer 1

man u most be on tensor and pde now...lol :)

Answer 2

on ... just for fun. if possible quick ... man

Answer 3

just this is bugging me so badly
Exercise 1.1.8.

Answer 4

https://web.williams.edu/go/math/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf

Answer 5

Ctrl + F ... search 1.1.8.
Don't have much time to waste

Answer 6

Actually, this question is very easy.. @experimentX

Answer 7

how ??

Answer 8

Did not you get the question clearly ??

It is saying to prove that thing..

Answer 9

how do you prove that?

Answer 10

Actually, I am talking about question..

Its solution is easy or not I don't know..

Ha ha ha..

Answer 11

already wasted 1 hrs ... can't waste another ... lol

Answer 12

Actually, that pdf is quite useful for me..

Thanks for giving me that..

Answer 13

some fool upvoted me :D ... that doesn't work at all .... lol

Answer 14

man second one's RHS will simplify to\[a^2+b^2+c^2\]

Answer 15

yeah ... i know that ... just need to prove it.

Answer 16

olypiad problems ... i find it extremely difficult even at my level man!! i need to dig a hole hide.

Answer 17

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}]

Answer 18

for the first one use AM_GM 4 times\[\frac{a^2}{b}+b\ge 2a\]\[\frac{b^2}{c}+c\ge 2b\]...

Answer 19

crazy crafty manipulation

Answer 20

the second one ... i can prove half of the original problem without it.

Answer 21

the last inequality i can get bu the use of quadratic formula ... but can't find any use of ... inequality n<=3

Answer 22

particularly I mean this
http://www.wolframalpha.com/input/?i=Minimize [{x+%2B+3+n%2Fx%2C++x+%3E%3D+3}%2C+{x}]

Answer 23

what kind of problems is this
x + 3n/x ... for x >= 3

Answer 24

sorry man i was out...how u realte this to second one?

Answer 25

can't ...have to do without it.

Answer 26

don't know what property is used here ..

Answer 27

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

Answer 28

which book do you use man??

Answer 29

oh sorry man original book is in persian and not translated to other languages :(

Answer 30

this is 500 pages about geometry and inequalities

Answer 31

you follow a very good writer!!

Answer 32

yeah :)

Answer 33

see u later santosh

Answer 34

see ya :) thanks for help man

Answer 35

Are you Linkha Santosh @experimentX ?? Sorry if there is any spelling mistake in name..

Answer 36

no mistake ... the first is my surname.

Answer 37

Santosh Linkha ..??

Answer 38

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..

Answer 39

lol ... what words?

Answer 40

I have written those..

"Everyone is on Facebook"..

I think these you said to Yahoo that day..

Answer 41

oh yeah ... kinda remembered on thing
"if you are searched in Google instead of fB then you are successful person"

Answer 42

Ha ha ha...

Yeah, this is true one..