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inequal...

by maihuna » Tue Dec 08, 2009 5:52 am
For positive x,y integer whether:
x/(y)^1/2 + y/(x)^1/2 >= (x)^1/2 + (y)^1/2

1. x^2 > y and y^2 > x
2. x, y are perfect squares
3. x>y
4. y>x
5. Any integer
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by mehravikas » Tue Dec 08, 2009 7:54 pm
I don't know whether I have understood the question correctly. Is it - c?
maihuna wrote:For positive x,y integer whether:
x/(y)^1/2 + y/(x)^1/2 >= (x)^1/2 + (y)^1/2

1. x^2 > y and y^2 > x
2. x, y are perfect squares
3. x>y
4. y>x
5. Any integer
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by aspirant1 » Tue Dec 08, 2009 10:22 pm
I got (E) - any integer

what's the IMO?
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by bou_k » Wed Dec 09, 2009 1:13 am
It should be E:
square both term in the equality so you get:
x^2/y +y^2/x +2 xy/sqrt(xy) >= x + y + 2 sqrt(xy)
and simplify to get
x^2/y +y^2/x >= x + y
x^2/y -x >= y- y^2/x
x(x/y-1) >= y(1-y/x)
x(x-y)/y>=y(x-y)/x
x^2(x-y)>=y^2(x-y)
x^2(x-y)-y^2(x-y)>=0
(x^2-y^2)(x-y)=(x-y)^2(x+1)>=0 this is always >=0 regardless of integers x and y then the solution must be E
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by gmatv09 » Wed Dec 09, 2009 3:21 pm
after simplification: x >= y (3)
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by satish.nagdev » Thu Dec 10, 2009 12:02 am
x>y for me too, i did this way

x/(y)1/2 + y/(x)^1/2 >= x^1/2 + y^1/2

x*x^1/2 + y*y^1/2 >=(xy)^1/2 * (x^1/2 + y^1/2)
>= x*(y^1/2) + y*(x^1/2)
x*x^1/2 - y*(x^1/2) >=x*(y^1/2) - y*y^1/2
x^1/2*(x-y) >= y^1/2 * (x-y)
x^1/2 >= y^1/2
x >=y
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by maihuna » Thu Dec 10, 2009 10:24 am
satish.nagdev wrote:x>y for me too, i did this way

x/(y)1/2 + y/(x)^1/2 >= x^1/2 + y^1/2

x*x^1/2 + y*y^1/2 >=(xy)^1/2 * (x^1/2 + y^1/2)
>= x*(y^1/2) + y*(x^1/2)
x*x^1/2 - y*(x^1/2) >=x*(y^1/2) - y*y^1/2
x^1/2*(x-y) >= y^1/2 * (x-y)
x^1/2 >= y^1/2
x >=y
But x=4 y=1 satisfy it right:
4/1+1/2 >= 2+1
or 4.5>=3

So x<y too satisfy the enquality, so either there is some derivation issue or ..?
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