Let a, b, and c be non-negative real numbers. Prove the following inequality:

- zepdrix

Let a, b, and c be non-negative real numbers. Prove the following inequality:

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- chestercat

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- zzr0ck3r

same here

- 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

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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## More answers

- misty1212

\[(x+b+c)^2\geq 0\] no matter what

- zepdrix

Oh haha good point XD

- dan815

how about thinking about it like the sum sub of 3 perfect square vs the sum of interchanging perfect squares

- dan815

so like rectangles formed from mixing and matching the sides of perfect squares

- dan815

|dw:1441246380649:dw|

- zzr0ck3r

Use Cauchy / Shwarz

- misty1212

i think the trick is to write
\[a^2+b^2+c^2-ab-bc-ab\] as a sum of squares

- dan815

|dw:1441246436682:dw|

- zepdrix

Doh >.<
Yah teacher mentioned something about Cauchy, I thought it was for a different problem though.. hmm

- zzr0ck3r

You are allowed?

- misty1212

think we can do it without that maybe

- dan815

hey but loook my picture works!!

- zepdrix

I am allowed to use Cauchy Schwarz if I first first prove it lol.
That's a separate problem though I suppose :)

- zepdrix

The squares thing? :O works?

- dan815

if you arrange any set of A^2,B^2,C^2 whjere A>B>C

- zzr0ck3r

it is not hard to prove for \(\mathbb{R}\).

- 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

Else, I would assume wlog a<=b<=c to get rid of so many cases.

- misty1212

since the right hand side is clearly positive, you get
\[a^2+b^2+c^2-ab-ac-bc\geq 0\]

- anonymous

you don't need cases, do what @misty1212 said, except the algebra is not quite right

- 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

it is a common gimmick is all i think

- zepdrix

Ooo that works out very nicely Misty!! :O

- zepdrix

Hmm I have a couple more like this :)
Imma give them a shot though before I bug you guys.

- zzr0ck3r

Another great trick in these is to add 0. like +a-a ...

- zepdrix

ya ya ya ya ya :) good call boss

- 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

hmm

- zzr0ck3r

Well, if you expand the RHS you get 8 terms, maybe use the last one?

- ganeshie8

suppose x,y,z are positive real numbers
can we say min(xyz) = min(x)*min(y)*min(z) ?

- zepdrix

ya you get 8 terms :d
two of which combine to give us some abc's.
Hmm.
min what? +_+

- 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

presuming \(a,b,c\) are positive reals
which keeps the sums \(a+b, ~b+c, ~c+a\) also positive

- zepdrix

Hmm neato :) Thanks

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