experimentX
  • experimentX
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 \( a+b+c=ab+bc+ca \) \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
man u most be on tensor and pde now...lol :)
experimentX
  • experimentX
on ... just for fun. if possible quick ... man
experimentX
  • experimentX
just this is bugging me so badly Exercise 1.1.8.

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More answers

experimentX
  • experimentX
https://web.williams.edu/go/math/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf
experimentX
  • experimentX
Ctrl + F ... search 1.1.8. Don't have much time to waste
anonymous
  • anonymous
Actually, this question is very easy.. @experimentX
experimentX
  • experimentX
how ??
anonymous
  • anonymous
Did not you get the question clearly ?? It is saying to prove that thing..
experimentX
  • experimentX
how do you prove that?
anonymous
  • anonymous
Actually, I am talking about question.. Its solution is easy or not I don't know.. Ha ha ha..
experimentX
  • experimentX
already wasted 1 hrs ... can't waste another ... lol
anonymous
  • anonymous
Actually, that pdf is quite useful for me.. Thanks for giving me that..
experimentX
  • experimentX
some fool upvoted me :D ... that doesn't work at all .... lol
anonymous
  • anonymous
man second one's RHS will simplify to\[a^2+b^2+c^2\]
experimentX
  • experimentX
yeah ... i know that ... just need to prove it.
experimentX
  • experimentX
olypiad problems ... i find it extremely difficult even at my level man!! i need to dig a hole hide.
experimentX
  • experimentX
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}]
anonymous
  • anonymous
for the first one use AM_GM 4 times\[\frac{a^2}{b}+b\ge 2a\]\[\frac{b^2}{c}+c\ge 2b\]...
experimentX
  • experimentX
crazy crafty manipulation
experimentX
  • experimentX
the second one ... i can prove half of the original problem without it.
experimentX
  • experimentX
the last inequality i can get bu the use of quadratic formula ... but can't find any use of ... inequality n<=3
experimentX
  • experimentX
particularly I mean this http://www.wolframalpha.com/input/?i=Minimize[{x+%2B+3+n%2Fx%2C++x+%3E%3D+3}%2C+{x}]
experimentX
  • experimentX
what kind of problems is this x + 3n/x ... for x >= 3
anonymous
  • anonymous
sorry man i was out...how u realte this to second one?
experimentX
  • experimentX
can't ...have to do without it.
experimentX
  • experimentX
don't know what property is used here ..
anonymous
  • anonymous
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\[\frac{ab^{3}}{c}+\frac{bc^{3}}{a}\ge 2b^{2}c\]...
experimentX
  • experimentX
which book do you use man??
anonymous
  • anonymous
oh sorry man original book is in persian and not translated to other languages :(
anonymous
  • anonymous
this is 500 pages about geometry and inequalities
experimentX
  • experimentX
you follow a very good writer!!
anonymous
  • anonymous
yeah :)
anonymous
  • anonymous
see u later santosh
experimentX
  • experimentX
see ya :) thanks for help man
anonymous
  • anonymous
Are you Linkha Santosh @experimentX ?? Sorry if there is any spelling mistake in name..
experimentX
  • experimentX
no mistake ... the first is my surname.
anonymous
  • anonymous
Santosh Linkha ..??
anonymous
  • anonymous
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..
experimentX
  • experimentX
lol ... what words?
anonymous
  • anonymous
I have written those.. "Everyone is on Facebook".. I think these you said to Yahoo that day..
experimentX
  • experimentX
oh yeah ... kinda remembered on thing "if you are searched in Google instead of fB then you are successful person"
anonymous
  • anonymous
Ha ha ha... Yeah, this is true one..

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