TEST For International Mathemathic Olympiad 2014

TEST for International Mathemathic Olympiad 2014

Breakfast of IMO Math Part 1

You may read this writing from notebook for
Convenience and completeness of a proof
1 ∀a,b,c>0, prove that
2((a+b)/2-√ab)≤3((a+b+c)/3-∛abc)

Prove that if 2-y-|x|=0, then x^2+y^2≥2

3 Give |x|>1,|y|>1, prove that
|a+b|≤|ab+1|

4 ∀x,y,z>0, prove that
√(xy+4yz+4zx)/(x+y)+√(yz+4xz+4xy)/(y+z)+√(xz+4xy+4yz)/(z+x)≥9/2

5 ∀a,b,c>0, prove that
1/(a^2+b^2 )+1/(b^2+c^2 )+1/(c^2+a^2 )≥27/(2(a+b+c)^2 )

GOOD LUCK

Solution :

Proof 1 for ∀a,b,c>0
We will prove by using well sharing method
Considering, 2((a+b)/2-√ab)≤3((a+b+c)/3-∛abc)
↔a+b-2√ab≤a+b+c-3∛abc
↔-2√ab≤c-3∛abc
↔ 3∛abc≤2√ab+c
Next we will show that
c/∛abc+√ab/∛abc+√ab/∛abc≥3
Considering, c/∛abc+√ab/∛abc+√ab/∛abc
=√(6&c^6/(a^2 b^2 c^2 ))+√(6&(a^3 b^3)/(a^2 b^2 c^2 ))+√(6&(a^3 b^3)/(a^2 b^2 c^2 ))
=√(6&c^4/(a^2 b^2 ))+√(6&ab/c^2 )+√(6&ab/c^2 )
≥√(6&c^4/(a^2 c^2 ))+√(6&ab/c)+√(6&ab/b^2 )
≥√(6&c^4/(c^2 c^2 ))+√(6&ab/a^2 )+√(6&ab/b^2 )
=√(6&1)+√(6&b/a)+√(6&a/b)
≥1+2=3 OK
Therefore, it is to be true.

2 Since 2-y-|x|=0 will obtain that
|x|+y=2
|x|^2+2|x|y+y^2=4
But |x|+y=2, then |x|y≤1
Then 2|x|y≤2
It imply that |x|^2+y^2≥2
That is, x^2+y^2≥2 OK

3 Give |x|>1,|y|>1, we have
x^2>1 and y^2>1
x^2-1>0 and y^2-1>0
x^2 (y^2-1)>y^2-1 and
y^2 (x^2-1)>x^2-1
Which it make x^2 y^2-x^2>y^2-1
And y^2 x^2-y^2>x^2-1
Next, x^2 y^2+1>x^2+y^2
x^2 y^2+2xy+1>x^2+2xy+y^2
(xy+1)^2>(x+y)^2
|xy+1|^2>|x+y|^2
Therefore, |xy+1|>|x+y| OK

4 For all x,y,z>0
We will prove by using well sharing method
Seeing, √(xy+4yz+4zx)/(x+y)+√(yz+4xz+4xy)/(y+z)+√(xz+4xy+4yz)/(z+x)
≥√(xy+4xy+4zx)/(x+y)+√(yz+4xz+4yz)/(y+z)+√(xz+4xy+4yz)/(z+x)
≥√(xy+4xy+4xy)/(x+y)+√(yz+4xz+4yz)/(y+z)+√(xz+4zx+4yz)/(z+x)
≥√(xy+4xy+4xy)/(x+y)+√(yz+4yz+4yz)/(y+z)+√(xz+4zx+4zx)/(z+x)
≥√9xy/(x+x)+√9yz/(y+z)+√9xz/(z+y)
≥(3√xy)/(x+x)+(3√yz)/(y+z)+(3√xz)/(z+y)
≥(3√xy)/(x+x)+(3√yz)/(y+y)+(3√xz)/(z+z)
=3/2 (√(xy/x^2 )+√(yz/y^2 )+√(zx/z^2 ))
=3/2 (√(y/x)+√(z/y)+√(x/z))
≥3/2 x3= 9/2 OKO
Therefore, it is to be true.

5 For all a,b,c>0
We will show by using well sharing method
1/(a^2+b^2 )+1/(b^2+c^2 )+1/(c^2+a^2 )≥27/(2(a+b+c)^2 )
↔(a+b+c)^2/(a^2+b^2 )+(a+b+c)^2/(b^2+c^2 )+(a+b+c)^2/(c^2+a^2 )≥27/2
We will show that it is to be true
Considering, (a+b+c)^2/(a^2+b^2 )+(a+b+c)^2/(b^2+c^2 )+(a+b+c)^2/(c^2+a^2 )
≥(a+a+c)^2/(a^2+b^2 )+(b+b+c)^2/(b^2+c^2 )+(a+b+c)^2/(c^2+a^2 )
≥(a+a+a)^2/(a^2+b^2 )+(b+b+c)^2/(b^2+c^2 )+(c+b+c)^2/(c^2+a^2 )
≥(a+a+a)^2/(a^2+b^2 )+(a+b+b)^2/(b^2+c^2 )+(c+c+c)^2/(c^2+a^2 )
≥(9a^2)/(a^2+a^2 )+(9b^2)/(b^2+c^2 )+〖9c〗^2/(c^2+b^2 )
≥(3a^2)/(a^2+a^2 )+(3b^2)/(b^2+b^2 )+〖3c〗^2/(c^2+c^2 )
=(3a^2)/(2a^2 )+(3b^2)/(2b^2 )+〖3c〗^2/(2c^2 )
=3/2+3/2+3/2=9/2
Therefore, 1/(a^2+b^2 )+1/(b^2+c^2 )+1/(c^2+a^2 )≥27/(2(a+b+c)^2 )

GOOD LUCK