zepdrix
  • zepdrix
Let a, b, and c be non-negative real numbers. Prove the following inequality:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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zzr0ck3r
  • zzr0ck3r
same here
zepdrix
  • zepdrix
\(\large\rm ab+bc+ca\le a^2+b^2+c^2\) 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
  • misty1212
idk but i think you are going to need something else, since you did not really use the fact that they are non negative

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misty1212
  • misty1212
\[(x+b+c)^2\geq 0\] no matter what
zepdrix
  • zepdrix
Oh haha good point XD
dan815
  • dan815
how about thinking about it like the sum sub of 3 perfect square vs the sum of interchanging perfect squares
dan815
  • dan815
so like rectangles formed from mixing and matching the sides of perfect squares
dan815
  • dan815
|dw:1441246380649:dw|
zzr0ck3r
  • zzr0ck3r
Use Cauchy / Shwarz
misty1212
  • misty1212
i think the trick is to write \[a^2+b^2+c^2-ab-bc-ab\] as a sum of squares
dan815
  • dan815
|dw:1441246436682:dw|
zepdrix
  • zepdrix
Doh >.< Yah teacher mentioned something about Cauchy, I thought it was for a different problem though.. hmm
zzr0ck3r
  • zzr0ck3r
You are allowed?
misty1212
  • misty1212
think we can do it without that maybe
dan815
  • dan815
hey but loook my picture works!!
zepdrix
  • zepdrix
I am allowed to use Cauchy Schwarz if I first first prove it lol. That's a separate problem though I suppose :)
zepdrix
  • zepdrix
The squares thing? :O works?
dan815
  • dan815
if you arrange any set of A^2,B^2,C^2 whjere A>B>C
zzr0ck3r
  • zzr0ck3r
it is not hard to prove for \(\mathbb{R}\).
misty1212
  • misty1212
\[a^2+b^2+c^2-ab-ac-bc=(a-b)^2+(b-c)^2+(c-a)^2\] or maybe half of that
zzr0ck3r
  • zzr0ck3r
Else, I would assume wlog a<=b<=c to get rid of so many cases.
misty1212
  • misty1212
since the right hand side is clearly positive, you get \[a^2+b^2+c^2-ab-ac-bc\geq 0\]
anonymous
  • anonymous
you don't need cases, do what @misty1212 said, except the algebra is not quite right
zepdrix
  • zepdrix
Oh 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
  • anonymous
it is a common gimmick is all i think
zepdrix
  • zepdrix
Ooo that works out very nicely Misty!! :O
zepdrix
  • zepdrix
Hmm I have a couple more like this :) Imma give them a shot though before I bug you guys.
zzr0ck3r
  • zzr0ck3r
Another great trick in these is to add 0. like +a-a ...
zepdrix
  • zepdrix
ya ya ya ya ya :) good call boss
zepdrix
  • zepdrix
I'll post the next one just in case anyone wanted to look at it :d\[\large\rm 8abc\le(a+b)(b+c)(c+a)\]
zepdrix
  • zepdrix
hmm
zzr0ck3r
  • zzr0ck3r
Well, if you expand the RHS you get 8 terms, maybe use the last one?
ganeshie8
  • ganeshie8
suppose x,y,z are positive real numbers can we say min(xyz) = min(x)*min(y)*min(z) ?
zepdrix
  • zepdrix
ya you get 8 terms :d two of which combine to give us some abc's. Hmm. min what? +_+
ganeshie8
  • ganeshie8
\[\min\{(a+b)(b+c)(c+a)\} =\min\{a+b)\} *\min\{b+c\}* \min\{c+a\} \] By AM-GM inequality we have \(a+b\ge 2\sqrt{ab}\) giving the desired result
ganeshie8
  • ganeshie8
presuming \(a,b,c\) are positive reals which keeps the sums \(a+b, ~b+c, ~c+a\) also positive
zepdrix
  • zepdrix
Hmm neato :) Thanks

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