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