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zepdrix
 one year ago
Let a, b, and c be nonnegative real numbers. Prove the following inequality:
zepdrix
 one year ago
Let a, b, and c be nonnegative real numbers. Prove the following inequality:

This Question is Closed

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm ab+bc+ca\le a^2+b^2+c^2\) So here is was I was attempting... For \(\large\rm a,~b,~c\ge0\) we have \(\large\rm (a+b+c)^2\ge0\). Expanding this out gives us: \(\large\rm a^2+b^2+c^2+2ab+2bc+2ca\ge0\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3idk but i think you are going to need something else, since you did not really use the fact that they are non negative

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3\[(x+b+c)^2\geq 0\] no matter what

dan815
 one year ago
Best ResponseYou've already chosen the best response.1how about thinking about it like the sum sub of 3 perfect square vs the sum of interchanging perfect squares

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so like rectangles formed from mixing and matching the sides of perfect squares

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3i think the trick is to write \[a^2+b^2+c^2abbcab\] as a sum of squares

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Doh >.< Yah teacher mentioned something about Cauchy, I thought it was for a different problem though.. hmm

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3think we can do it without that maybe

dan815
 one year ago
Best ResponseYou've already chosen the best response.1hey but loook my picture works!!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I am allowed to use Cauchy Schwarz if I first first prove it lol. That's a separate problem though I suppose :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The squares thing? :O works?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if you arrange any set of A^2,B^2,C^2 whjere A>B>C

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3it is not hard to prove for \(\mathbb{R}\).

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3\[a^2+b^2+c^2abacbc=(ab)^2+(bc)^2+(ca)^2\] or maybe half of that

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Else, I would assume wlog a<=b<=c to get rid of so many cases.

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3since the right hand side is clearly positive, you get \[a^2+b^2+c^2abacbc\geq 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you don't need cases, do what @misty1212 said, except the algebra is not quite right

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh so the idea I started with was kind of a mess? Where do you come up with the subtraction like that though? :d hmm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is a common gimmick is all i think

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ooo that works out very nicely Misty!! :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Hmm I have a couple more like this :) Imma give them a shot though before I bug you guys.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Another great trick in these is to add 0. like +aa ...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2ya ya ya ya ya :) good call boss

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I'll post the next one just in case anyone wanted to look at it :d\[\large\rm 8abc\le(a+b)(b+c)(c+a)\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Well, if you expand the RHS you get 8 terms, maybe use the last one?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0suppose x,y,z are positive real numbers can we say min(xyz) = min(x)*min(y)*min(z) ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2ya you get 8 terms :d two of which combine to give us some abc's. Hmm. min what? +_+

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\min\{(a+b)(b+c)(c+a)\} =\min\{a+b)\} *\min\{b+c\}* \min\{c+a\} \] By AMGM inequality we have \(a+b\ge 2\sqrt{ab}\) giving the desired result

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0presuming \(a,b,c\) are positive reals which keeps the sums \(a+b, ~b+c, ~c+a\) also positive
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