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Topic: Inequality / Sat II Physics Book (Read 3238 times) |
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apsurf1729
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Inequality / Sat II Physics Book
« on: Feb 26th, 2006, 2:31pm » |
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IMO ( Columbia 2001) Prove the inequality
(x, y element R)
3(x+y+1) ^2 +1 greater than or equal to 3xy
I tried to use the trick with AM-GM but it is not working for me
Could someone solve this problem and show me how to do it
Also does anyone know if Barrens Study guide for the Physics Sat II is actually representative of it, if not what are some good books to study for the Sat II physics
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Icarus
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Re: Inequality / Sat II Physics Book
« Reply #1 on: Feb 26th, 2006, 3:08pm » |
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I don't know if the IMO allows calculus, but one way is to treat y as constant, and consider the following function of x: f(x) = 3(x+y+1)2 + 1 - 3xy.
set 0 = f'(x) = 6(x+y+1) - 3y. So f has an extremal point at x = -1 - y/2. You can see that this extrema is a minimum by either noting that f''(-1-y/2) = 6 > 0, or by noting that f --> oo as x --> +/- oo.
f(-1-y/2) = (3y2 + 6y + 16)/4. The minimum value of this parabola is 13/4 > 0.
Since the minimum value of f(x) > 0 for all y, 3(x+y+1)2 + 1 > 3xy for all x, y.
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apsurf1729
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Re: Inequality / Sat II Physics Book
« Reply #2 on: Feb 26th, 2006, 5:44pm » |
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Is there any algebraic way like using AM-GM Inequality, Cauchy-Schwarz or something along those lines that would allow for a substitution
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Eigenray
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Re: Inequality / Sat II Physics Book
« Reply #3 on: Feb 27th, 2006, 1:13am » |
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Calculus can be avoided by completing the square:
3(x+y+1)2+1-3xy = 3(x2+(y+2)x) + 3y2+6y+4
= 3(x+(y+2)/2)2 + 3y2+6y+4 - 3(y+2)2/4
= 3(x+(y+2)/2)2 + (9y2 + 12y + 4)/4
= 1/4 [ 3(2x+y+2)2 + (3y+2)2 ] > 0.
But this is not very symmetric. Another way to look at it is to put s=x+y, p=xy, so that we are to prove
3(s+1)2 + 1 - 3p > 0.
Now, s,p can be any real numbers subject to
s2 - 4p = (x-y)2 > 0,
so it makes sense to handle the -3p term by writing
3(s+1)2 + 1 - 3p = 3/4(s2-4p) + (9s2 + 24s + 16)/4
= 1/4 [ 3(s2-4p) + (3s+4)2 ]
= 1/4 [ 3(x-y)2 + (3x+3y+4)2 ]
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