Next 10 problems aren’t mine, but they belong to my Friends who have magnificent wisdom, may I must Admire all, although I will not state names. May you 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) 2 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 ) 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) Proof 1for ∀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| OKK 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 ) 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
Posted on: Mon, 13 Oct 2014 09:10:10 +0000
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