Breakfast of IMO Math Part 2

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.

6 ∀a,b,c>0, prove that
abc≥(a+b-c)(b+c-a)(c+a-b)

7 Let a,b,c>0, such that
a^2 b^2+b^2 c^2+c^2 a^2=3 Prove that
(a+b)(c^2+1/a^2 )+(b+c)(a^2+1/b^2 )+
(c+a)(b^2+1/c^2 )≥2abc+6

8 ∀a,b,c>0,a+b+c=3 prove that
1+8acb≥9min{a,b,c}

9 Let x,y,and z be consecutive Prime
Prove that x+y≥z

10 Given a,b,c be positive integers such that
a^2/(b+c),b^2/(c+a),c^2/(a+b) be positive integers
Prove (a,b,c)>1 (G.C.D)

Solution :

6 For all a,b,c>0
We will show by using well sharing method
Considering, (a+b-c)(b+c-a)
= ab+ac-a^2+b^2+bc-ba-cb-c^2+ca
=b^2+2ac-a^2-c^2
≤b^2+2ac-ac-ac
=b^2
Next b^2 (c+a-b)
=b(bc+ba-b^2 )
≤b(ac+bb-b^2 )
= b(ac+b^2-b^2 )
=abc
Therefore, abc≥(a+b-c)(b+c-a)(c+a-b)
7 Let a,b,c>0, such that
a^2 b^2+b^2 c^2+c^2 a^2=3abc
So, ab/c+bc/a+ca/b=3, by using well sharing method
≥ab/a+bc/c+ca/b
≥ab/a+bc/a+ca/c
=a+b+c
Which, make a^2+b^2+c^2≤3
1/a^2 +1/b^2 +1/c^2 ≥3
But, (a+b)(c^2+1/a^2 )+(b+c)(a^2+1/b^2 )+
+(c+a)(b^2+1/c^2 )≥(a+b)(a^2+1/a^2 )+
+(b+c)(c^2+1/b^2 )+(c+a)(b^2+1/c^2 )
≥(a+b)(a^2+1/a^2 )+
+(b+c)(b^2+1/b^2 )+(c+a)(c^2+1/c^2 )
≥(a+b)2+(b+c)2+(c+a)2
≥2(a+b+c)+2(a+b+c)
Hence,(a+b)(c^2+1/a^2 )+(b+c)(a^2+1/b^2 )+
+(c+a)(b^2+1/c^2 )≥2(a+b+c)+6
Because 2(a+b+c)≤6 OKO (it abstract)

8 ∀a,b,c>0,a+b+c=3, we have
0<abc≤1, 0<min{a,b,c}≤1,
And as a≤abc,if a≤b≤c
So, min{a,b,c}≤abc
9min{a,b,c}≤9abc
≤8abc+1

9 Let x,y,and z be consecutive Prime
From the proof of Pual Erdos say that
All positive integers n are greater than 1
From n to 2n there exist an integer k
Being Prime
In this case if assume that x+y<z
Then will make the proof of Pual Erdos may
Has a contradiction is that z≥2y
That is, among y and 2y has no Prime
Such as x=17,y=19,z is ?
As x+y<z,so z=37
Because x+y=17+19=36
There is a contradiction is that
Among 17 and 34 has no Prime OK

10 Given a,b,b be positive integers such that
a^2/(b+c),b^2/(c+a),c^2/(a+b) be positive integers
Hence, ((b+c))⁄a^2 ,((c+a))⁄b^2 ,((a+b))⁄c^2
a^2=k(b+c),b^2=m(c+a),c^2=n(a+b)
a=√(k(b+c) ),b=√(m(c+a) ),c=√(n(a+b) )
If give (a,b,c)=d, then
(√(k(b+c) ),√(m(c+a) ),√(n(a+b) ))=d
From its definition, we have
d⁄√(k(b+c) ),d⁄√(m(c+a) ),d⁄√(n(a+b) )
dx=√(k(b+c) ),dy=√(m(c+a) ) and
dz=√(n(a+b) ) Which imply that
d^2 x^2= k(b+c), d^2 y^2= m(c+a), and
d^2 z^2= m(a+b)
But, (b+c)∤x^2,(c+a)∤y^2,(a+b)∤z^2
Thence, ((b+c))⁄d^2 ,((c+a))⁄d^2 ,((a+b))⁄d^2
We see that d>1 because b+c,c+a,a+b≥2
Therefore, its G. C.D must be greater than 1
Acknowledgement