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This time I need to prove the following:
a*b + a*c + b*c <= a^2 + b^2 + c^2
Any ideas how that can be done?
Well you might imagine a, b, and c to be the sides of a cube with rectangular sides.
Then a*b, a*c, and b*c are the areas of three sides of the cube.
Next, I don't know what comes next...
igloo myrtilles fourmis
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This time I need to prove the following:
a*b + a*c + b*c <= a^2 + b^2 + c^2Any ideas how that can be done?
well..... lets see
given that a>0, b>0, c>0
The Beginning Of All Things To End.
The End Of All Things To Come.
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Wait, but what does that give us?
Wait, but what does that give us?
nothing that was just me trying to get it
The Beginning Of All Things To End.
The End Of All Things To Come.
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I got it!!
Make a, b, and c these names: M for middle #. M + x for large number. M - y for small number.
Now x and y are greater than or equal to zero.
Substitute in and expand and cancel stuff and you get
the result that the big side of the equation is bigger by x^2 + xy + y^2.
No time to explain. But seems to be a positive amount bigger.
igloo myrtilles fourmis
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This time I need to prove the following:
a*b + a*c + b*c <= a^2 + b^2 + c^2Any ideas how that can be done?
I got it!!
Make a, b, and c these names: M for middle #. M + x for large number. M - y for small number.
Now x and y are greater than or equal to zero.
Substitute in and expand and cancel stuff and you get
the result that the big side of the equation is bigger by x^2 + xy + y^2.
No time to explain. But seems to be a positive amount bigger.
ok, by that
since x,y >= 0, -xy will ALWAYS be negative
and even if not, x^2 + y^2 will ALWAYS be positive (unless there 0, in which case 0 is <= 0 anyways)
therefore
a*b + a*c + b*c <= a^2 + b^2 + c^2
goodjob frank
(rewritten with math tags)
Last edited by luca-deltodesco (2006-06-12 10:17:20)
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Great job! Franklin!
X'(y-Xβ)=0
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Thanks Frank!
You're brilliant!
Thanks for the thank you's! Just sorry I don't explain myself better, but luca did a nice job doing that!
igloo myrtilles fourmis
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