Next 15 points aren’t mine but they belong to my Friends which may I praise them as courageous men and my idol 1 give f:N→R Defined by f(f(x) )=(2f(x)-2)/(f(x)) Find f Solution we must thank Swiss mathematician Euler who Has assigned symbol y=f(x) which we can interpret As follows f(x)→f(A) arise from substitution x by A,A∈D_f f(A)→f(x) arise from substitution A by x,x∈D_f And when R_f∩D_f≠∅, we have fof(x)=f(f(x) ),x∈D_fof, where D_fof={x∈D_f│f(x)∈D_f } We assume that there exist x,and f(x)∈D_f Hence, we have f(x)=(2x-2)/x Recheck solution f(f(x))=(2((2x-2)/x)-2)/(((2x-2)/x) ) =(x-2)/(x-1) =((2/(2-f(x)))-2)/((2/(2-(x)))-1) =(2f(x)-2)/(f(x)): ok From above, we see D_f=N,R_f={(2x-2)/x│x∈N} And N∩{ (2x-2)/x│x∈N}≠∅ And D_fof={x∈D_f│f(x)∈D_f } ={x∈N│(2x-2)/x∈N} ={2} Since, f(2)=1,and f(1)=0 So, fof(2)=(2-2)/(2-1)=0 Therefore, fof={(2,0) }:(2,1),(1,0)∈f If there is someone see the other function, please Share for our friends. 2 I have to thank the word Co-prime since I used not to Hear it before and used to hear only Relatively Prime. Give a,b,and c be positive integers which all two Numbers are Co-prime such that (a,b+c)>1,(b,c+a)>1,and (c,a+b)>1 Find the sum of a,b,c when a,b,c are the least Solution 1 as (a,b)=(a,a+b)=(b,a+b) And 2 [a,b](a,b)=|ab| So, (a,b+c)=(a,a+b+c) (b,c+a)=(b,a+b+c) (c,a+b)=(c,a+b+c) And, [a,a+b+c](a,a+b+c)=a(a+b+c) [b,a+b+c](b,a+b+c)=b(a+b+c) [c,a+b+c](c,a+b+c)=c(a+b+c) Hence, a+b+c=[a,a+b+c](a,a+b+c)/((a+b+c) )+ +[b,a+b+c](b,a+b+c)/((a+b+c) )+[c,a+b+c](c,a+b+c)/((a+b+c) ) And since, (a,a+b+c)>1, (b,a+b+c)>1,(c,a+b+c)>1, result in If (a,a+b+c)=2, then a=2m,∃m∈N If (b,a+b+c)=3, then b=3n,∃n∈N If (c,a+b+c)=5, then c=5k,∃k∈N, where (2m,2n)=(2n,2k)=(2k,2m)=1, where (m,n)=(n,k)=(k,m)=1 From clever selecting m,n,and k, we have a+b+c=30 =2+3+25 or =5+9+16 or =3+5+22 Remark: (b,a+b+c)≠2 because (a,b)=1 And (c,a+b+c)≠4 because (a,c)=1 3 give a,b,c>1, ab+bc+ca=2abc Show that √(a/(a-1))+√(b/(b-1))+√(c/(c-1))≤3√3 Solution as ab+bc+ca=2abc So, 1/a+1/b+1/c=2 And it make a+b+c≥9/2 And, 0≤2(a+b+c)-9 And, a+b+c≤3(a+b+c)-9, we have a≤3(a-1),or b≤3(b-1),or c≤3(c-1) Hence, a/(a-1)≤3,or b/(b-1)≤3,or c/(c-1)≤3 And, a/(a-1)+ b/(b-1)+c/(c-1)≤3+3+3 And, √(a/(a-1))+√(b/(b-1))+√(c/(c-1))≤√3+√3+√3 Therefore, √(a/(a-1))+√(b/(b-1))+√(c/(c-1))≤3√3 4 give xyz=1,and x,y,z>0, prove that x^2/(y+zx)+y^2/(z+xy)+z^2/(x+yz)≥3/2 Proof as if xyz=1,then x+y+z≥3 Then x^2+y^2+z^2≥x+y+z But, since, x^2+y^2+z^2≥xy+yz+zx Hence,2(x^2+y^2+z^2) ≥xy+yz+zx+x+y+z And we can write 2x^2≥y+zx,or 2y^2≥z+xy,or 2z^2≥x+yz And, x^2/(y+zx)≥1/2,or y^2/(z+xy)≥1/2,or z^2/(x+yz)≥1/2 That is, x^2/(y+zx)+y^2/(z+xy)+z^2/(x+yz)≥3/2 5 give a,b,c>0,and a+b+c=9 Prove that (a+1)/(b^2+1)+(b+1)/(c^2+1)+(c+1)/(a^2+1)≤6/5 Proof as a+b+c=9, we have a^2+b^2+c^2≥27 2(a^2+b^2+c^2)≥54 2(a^2+b^2+c^2 )+6= 2(a^2+1)+2(b^2+1)+2(c^2+1)≥60 But 5a+5b+5c+25= 5(a+1)+5(b+1)+5(c+1)=60 So, 5(a+1)+5(b+1)+5(c+1)≤ 2(a^2+1)+2(b^2+1)+2(c^2+1) And result in 5(a+1)≤2(b^2+1) or 5(b+1)≤2(c^2+1),5(c+1)≤2(a^2+1) And (a+1)/(b^2+1),(b+1)/(c^2+1),(c+1)/(a^2+1)≤2/5 Therefore, (a+1)/(b^2+1)+ (b+1)/(c^2+1)+(c+1)/(a^2+1)≤6/5 6 give a,b,c>0,a+b+c=3, prove that (a^2 b+b^2 c+c^2 a)(ab+bc+ca)≤9 Proof as (a+b+c)^3=27 and (a+b+c)^2=9 But a+b+c=3,we have It result in a^2 b+b^2 c+c^2 a≤3 and ab+bc+ca≤3 That is, (a^2 b+b^2 c+c^2 a)(ab+bc+ca)≤9 7 give x,y,z>0,x+y+z=xyz, prove that √(x^2/yz+2)+√(y^2/zx+2)+√(z^2/xy+2)≥xyz Proof as √(x^2+2yz)≥2√2 x√yz>x√yz And √(y^2+2zx)≥2√2 y√zx>y√zx And √(z^2+2xy)≥2√2 z√xy>z√xy Hence, √(x^2/yz+2)≥x,√(y^2/zx+2)≥y,and √(z^2/xy+2)≥z Therefore, √(x^2/yz+2)+√(y^2/zx+2)+√(z^2/xy+2) ≥x+y+x=xyz Remark: x=1,y=2,and z=3 8 give a,b,c>0,and abc=1, prove that 1/(a^2+2b^2+3)+1/(b^2+2c^2+3)+1/(c^2+2a^2+3)≤1/2 Proof as abc=1,we have a+b+c≥3 a^2+b^2+c^2≥3 2(a^2+b^2+c^2)≥6 So, a^2+2b^2+3+b^2+2c^2+3+ c^2+2a^2+3≥18=6+6+6 And we have a^2+2b^2+3≥6 or b^2+2c^2+3≥6,orc^2+2a^2+3≥6 And 1/(a^2+2b^2+3),1/(b^2+2c^2+3),and 1/(c^2+2a^2+3)≤1/6 Therefore, 1/(a^2+2b^2+3)+1/(b^2+2c^2+3)+1/(c^2+2a^2+3)≤1/2 9 give a,b,c>1,a+b+c=4, prove that 1/(a-1)+1/(b-1)+1/(c-1)≥8/(a+b)+8/(b+c)+8/(c+a) Proof Colloraly Let a,b,c be positive real We have 1 (a+b)(b+c)(c+a)≥8abc 2 √(a+b)+√(b+c)+√(c+a)≥√2a+√2b+√2c 3 1/(a+b)+1/(b+c)+1/(c+a)≤1/2a+1/2b+1/2c 4 1/√(a+b)+1/√(b+c)+1/√(c+a)≤1/√2a+1/√2b+1/√2c Proof are left as exercises. Remark: (a+b)+(b+c)+(c+a) =2a+ab+2c And if a≤b≤c,we have (b+c)-(a+b)=c-a ≤2c-2a=2(c-a) As 8/(a+b)+8/(b+c)+8/(c+a)=8/(4-c)+8/(4-a)+8/(4-b) So, we must show that 1/(a-1)+1/(b-1)+1/(c-1)≥8/(4-c)+8/(4-a)+8/(4-b) Seeing, 1=1 And 3/2-(a+b+c)/8=a+b+c-3 And (4-a)/8+(4-b)/8+(4-c)/8=(a-1)+(b-1)+(c-1) From Colloraly above, will get 8/(4-a)+8/(4-b)+8/(4-c)≤1/(a-1)+1/(b-1)+1/(c-1) Therefore, 1/(a-1)+1/(b-1)+1/(c-1)≥8/(a+b)+8/(b+c)+8/(c+a) 10 give a,b,c>0,abc=1 prove that (a+1/b)^2+(b+1/c)^2+(c+1/a)^2≥ ≥3(a+b+c+1) Proof (a+1/b)^2+(b+1/c)^2+(c+1/a)^2= a^2+2a/b+1/b^2 +b^2+2b/c+1/c^2 +c^2+2c/a+1/a^2 And since, abc=1 we obtain 1/a^2 +1/b^2 +1/c^2 ≥3,a/b+b/c+c/a≥a+b+c And a^2+b^2+c≥a+b+c Hence, a^2+2a/b+1/b^2 +b^2+2b/c+1/c^2 +c^2+2c/a+1/a^2 ≥3(a+b+c+1) That is, (a+1/b)^2+(b+1/c)^2+(c+1/a)^2≥ 3(a+b+c+1) 11 give a,b,c>0 prove that a/√(a^2+8bc)+b/√(b^2+8ca)+c/√(c^2+8ab)≥1 Proof as 9a^2+9b^2+9c^2≥ ≥a^2+8bc+b^2+8ca+c^2+8ab So, we can write 9a^2≥a^2+8bc or 9b^2≥b^2+8ca or 9c^2≥c^2+8ab So, a^2/(a^2+8bc )≥1/9 or b^2/(b^2+8ca)≥1/9 or or c^2/(c^2+8ab)≥1/9 So, a/√(a^2+8bc)≥1/3 or b/√(b^2+8ca)≥1/3 or c/√(c^2+8ab)≥1/3 Hence, a/√(a^2+8bc)+b/√(b^2+8ca)+c/√(c^2+8ab)≥1 Or Proof seeing, bc/a^2 +ca/b^2 +ab/c^2 ≥1+1+1 And 8bc/a^2 +8ca/b^2 +8ab/c^2 ≥8+8+8 And (a^2+8bc)/a^2 +(b^2+8ca)/b^2 +(c^2+8ab)/c^2 ≥9+9+9 And a^2/(a^2+8bc)+b^2/(b^2+8ca)+c^2/(c^2+8ab)≥1/9+1/9+1/9 And a/√(a^2+8bc)+b/√(b^2+8ca)+c/√(c^2+8ab)≥1/3+1/3+1/3 Therefore, a/√(a^2+8bc)+b/√(b^2+8ca)+c/√(c^2+8ab)≥1 12 give x,y,z>0,xyz=1 prove that x/(2yz+2)+y/(2xz+y)+z/(2xy+z)≥1 Proof as xyz=1,we will obtain 1/x^2 +1/y^2 +1/z^2 ≥3 And 2/x^2 +2/y^2 +2/z^2 ≥6 And (2+x^2)/x^2 +(2+y^2)/y^2 +(2+z^2)/z^2 ≥3+3+3 And x^2/(2+x^2 )+y^2/(2+y^2 )+z^2/(2+z^2 )≥1/3+1/3+1/3=1 And x/((2+x^2)/x)+y/((2+y^2)/y)+z/((2+z^2)/z)≥1 And x/(2/x+x)+y/(2/y+y)+z/(2/z+z)≥1 But xyz=1→yz=1/x,zx=1/y and xy=1/z Therefore, x/(2yz+2)+y/(2xz+y)+z/(2xy+z)≥1 ok 13 give a,b,c>0,ab+bc+ca=3abc, prove √ab/(√(3b+c)-√c)+√bc/(√(3c+a)-√a)+√ca/(√(3a+b)-√b)≤3√abc Proof as ab+bc+ca=3abc Hence, 1/a+1/b+1/c=3,a+b+c≥3 And 3b-2+3c-2+3a-2≥3 So, 3b-2≥ 1/c or 3c-2≥ 1/a or 3a-2≥ 1/b So, 3bc-2c≥1or 3ca-2a≥1or 3ab-2b≥1 So, 3bc+c^2≥(1+c)^2 or Or 3ca+a^2≥(1+a)^2 or 3ab+b^2≥(1+b)^2 So, √(3bc+c^2 )≥1+c or √(3ca+a^2 )≥1+a Or √(3ab+b^2 )≥1+b So, 1/(√(3b+c)-√c)≤√c or 1/(√(3c+a)-√a)≤√a or Or 1/(√(3a+b)-√b)≤√b So, √ab/(√(3b+c)-√c)≤√abc or √bc/(√(3c+a)-√a)≤√abc or Or √ca/(√(3c+a)-√a)≤√abc Hence, √ab/(√(3b+c)-√c)+√bc/(√(3c+a)-√a)+√ca/(√(3c+a)-√a) ≤3√abc 14 give a,b,c>0,abc=1 prove that a/√(b^2+2c)+b/√(c^2+2a)+c/√(a^2+2b)≥√3 Proof as abc=1→a+b+c≥3 →a^2+b^2+c^2≥a+b+c →3a^2+3b^2+〖3c〗^2≥ ≥b^2+2c+c^2+2a+a^2+2b So, 3a^2≥b^2+2c or 3b^2≥c^2+2a or Or 〖3c〗^2≥a^2+2b So, a^2/(b^2+2c )≥1/3 or b^2/(c^2+2a )≥1/3 or c^2/(a^2+2b )≥1/3 So, a/√(b^2+2c)≥1/√3 or b/√(c^2+2a)≥1/√3 or Or c/√(a^2+2b)≥1/√3 Hence, a/√(b^2+2c)+b/√(c^2+2a)+c/√(a^2+2b)≥√3 15 give x,y,z>0, prove that x/√((x+y)(x+z) )+y/√((y+x)(+z) )+z/√((z+x)(z+y) )≤3/2 Proof as a+b+c≤1,then abc≤1/27 And since, 8xyz≤x^2 y+x^2 z+y^2 x+y^2 z+ +z^2 x+z^2 y+2xyz,x,y,z>0 So, ((2x)(2y)(2z))/((x+y)(y+z)(z+x))= =((2x)(2y)(2z))/(√((x+y)(x+z)) √((y+x)(y+z)) √((z+x)(z+y)))≤1 So, ((2x)(2y)(2z))/(27√((x+y)(x+z)) √((y+x)(y+z)) √((z+x)(z+y)))≤1/27 So, 2x/(3√((x+y)(x+z)))+2y/(3√((y+x)(y+z)))+2z/(3√((z+x)(z+y))) ≤1/3+1/3+1/3=1 So, 2x/√((x+y)(x+z))+2y/√((y+x)(y+z))+2z/√((z+x)(z+y))≤ ≤3 Hence, x/√((x+y)(x+z))+y/√((y+x)(y+z))+z/√((z+x)(z+y)) ≤3/2 ok This writing, if there is a mistake, and then it is mine But if there is some profit that can make wisdom, then I Assign this success with Pro.Dr. Narong Phannim Who be my great teacher. Remark: if we think that varied problems are magnificent Food, then we will be capable to solve them happily And important we will get new wisdom by ourselves
Posted on: Tue, 27 Jan 2015 05:17:52 +0000
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