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same here

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

Oh haha good point XD

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

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Use Cauchy / Shwarz

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

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You are allowed?

think we can do it without that maybe

hey but loook my picture works!!

The squares thing? :O works?

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

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

\[a^2+b^2+c^2-ab-ac-bc=(a-b)^2+(b-c)^2+(c-a)^2\] or maybe half of that

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

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

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

it is a common gimmick is all i think

Ooo that works out very nicely Misty!! :O

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

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

ya ya ya ya ya :) good call boss

hmm

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

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

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

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

Hmm neato :) Thanks