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. For the following proofs are mine which may be True or false and the method that are helped to Prove in problem 1, 3, 4, 5, 6, and 7, if there are Profit, then may i assign its name “ The Sharing Method” 1 For all a,b,c∈R^+, prove that a/b+b/c+c/a+3(∛abc)/(a+b+c)≤4 2 For all a,b,c≥0, prove that a/(a+1)+b/(b+1)+c/(c+1)+1/(a+b+c+1)≥1 3 For all a,b,c∈R^+, prove that a^3+b^3+c^3+ab^2+bc^2+ca^2≥2(a^2 b+b^2 c+c^2 a) 4 For all x,y,z∈R^+, prove that x^2/y^2 +y^2/z^2 +z^2/x^2 ≥x/y+y/z+z/x 5 For all a,b,c,>0, prove that a/(a^2+bc+b^2 )+b/(b^2+ca+c^2 )+c/(c^2+ab+a^2 )≤(a+b+c)/(ab+bc+ca) 6 For all x,y,z≥0, prove that √2 (x+y+z)≤√(x^2+y^2 )+√(y^2+z^2 )+√(z^2+x^2 ) 7 For all a,b,c≥0, prove that 1/(a+b)+1/(b+c)+1/(c+a)+8/(a+b+c)≤6/√(ab+bc+ca) For all a,b,c∈R^+, and abc=1, prove that a^2+b^2+c^2≥a+b+c 8 For all x,y,z≥0 and x+y+z=3 prove that x^2 y+y^2 z+z^2 x+xyz≤4 9 For all a,b,c,d∈R, a+b+c+d=0 Prove that a^3+b^3+c^3+d^3= abc+bcd+cda+dab 10 Give {a_n } be a sequence in pattern {a_n }:3,7,13,21,31,… Prove that〖 a〗_i 〖.a〗_((i+1) )∈{a_n } Proof 1 As a,b,c∈R^+, we have a/b+b/c+c/a+3(∛abc)/(a+b+c) =(a^2 c(a+b+c)+b^2 a(a+b+c)+c^2 b(a+b+c)+3abc(∛abc))/abc(a+b+c) =(a^3 c+a^2 bc+a^2 c^2+a^2 b^2+b^3 a+ab^2 c+〖abc〗^2+b^2 c^2+c^3 b+3abc(∛abc))/abc(a+b+c) And as a^2 c^2+a^2 b^2+b^2 c^2+3abc(∛abc)≥ a^2 c^2+a^2 b^2+b^2 ab+abc(∛(c^2 c))+2abc(∛abc)≥ a^2 c^2+a^2 b^2+b^3 a+abc^2+2abc(∛abc)≥… Lastly, it must be greater or equal than c^3 b+a^3 c+b^3 a+a^2 bc+ab^2 c+abc^2 We will show that a^3 c+a^2 bc+b^3 a+〖ab〗^2 c+〖abc〗^2+c^3 b +c^3 b+c^3 b+a^3 c+b^3 a+a^2 bc+ab^2 c+abc^2 ≥4(a^2 bc+ab^2 c+abc^2 ) As a^3 c+c^3 b+b^3 a≥a^2 bc+ab^2 c+abc^2 Hence, a^3 c+a^2 bc+b^3 a+〖ab〗^2 c+〖abc〗^2+c^3 b+ +c^3 b+c^3 b+a^3 c+b^3 a+a^2 bc+ab^2 c+abc^2 ≥4(a^2 bc+ab^2 c+abc^2 ) That is, a/b+b/c+c/a+3(∛abc)/(a+b+c)≥4 2 from a,b,c≥0,we have 4 cases are 1 a=b=c=0 2 a=0,b,c≠0 3 a=b=0,c≠0 4 a,b,c≠0 But we will show in general form As 1/(a+b+c+1)=1-(a+b+c)/(a+b+c+1) Thence, a/(a+1)+b/(b+1)+c/(c+1)+1/(a+b+c+1) =1+a/(a+1)-a/(a+b+c+1)+b/(b+1)-b/(a+b+c+1)+c/(c+1)-c/(a+b+c+1) As a/(a+1)-a/(a+b+c+1),b/(b+1)-b/(a+b+c+1),c/(c+1)-c/(a+b+c+1)≥0 Therefore, a/(a+1)+b/(b+1)+c/(c+1)+1/(a+b+c+1)≥1 3 For all a,b,c∈R^+, will get that a^3+b^3+c^3+ab^2+bc^2+ca^2≥ =a^2 a+b^2 b+c^2 c+abb+bcc+caa ≥a^2 b+b^2 b+c^2 c+aab+bcc+caa ≥a^2 b+b^2 c+c^2 c+a^2 b+bbc+caa ≥a^2 b+b^2 c+c^2 a+a^2 b+b^2 c+cca =a^2 b+b^2 c+c^2 a+a^2 b+b^2 c+c^2 a Therefore, a^3+b^3+c^3+ab^2+bc^2+ca^2 ≥2(ab^2+bc^2+ca^2 ) 4 For all x,y,z∈R^+, we have x^2/y^2 +y^2/z^2 +z^2/x^2 =xx/yy+yy/zz+zz/xx ≥xy/yy+xy/zz+zz/xx ≥x/y+xy/zz+zz/xx =x/y+y/z+z/x Therefore, x^2/y^2 +y^2/z^2 +z^2/x^2 ≥x/y+y/z+z/x 5 For all a,b,c,d>0, we have a/(a^2+bc+b^2 )+b/(b^2+ca+c^2 )+c/(c^2+ab+a^2 ) =a/(aa+bc+bb)+b/(bb+ca+cc)+c/(cc+ab+aa) ≥a/(ab+bc+bb)+b/(ab+ca+cc)+c/(cc+ab+aa) ≥a/(ab+bc+bc)+b/(ab+ca+cc)+c/(bc+ab+aa) ≥a/(ab+bc+ca)+b/(ab+bc+cc)+c/(bc+ab+aa) ≥a/(ab+bc+ca)+b/(ab+bc+ca)+c/(bc+ab+ca) =(a+b+c)/(ab+bc+ca) Therefore, a/(a^2+bc+b^2 )+b/(b^2+ca+c^2 )+c/(c^2+ab+a^2 ) ≥(a+b+c)/(ab+bc+ca) 6 For all x,y,z≥0 will get that 1 x=y=z=0 2 x=0,y,z≠0 3 x=y=0,z≠0 4 x,y,z≠0 1and 3 are obvious 2 √(x^2+y^2 )+√(y^2+z^2 )+√(z^2+x^2 )= √(y^2 )+√(y^2+z^2 )+√(z^2 )= y+z+√(y^2+z^2 )≥√2 (y+z) As (y+z)/(y+z)+ √(y^2+z^2 )/(y+z)=1+√(y^2+z^2 )/(y+z)≥1+ 1/2≥√2 4 √(x^2+y^2 )+√(y^2+z^2 )+√(z^2+x^2 )≥ ≥√(x^2+x^2 )+√(y^2+y^2 )+√(z^2+z^2 ) √2 (x+y+x) OK From 4 cases it is to be true. 7 For all a,b,c∈R^+, and abc=1 We give a=x/y,b=y/z,and c=z/x, we have abc=1↔(abc)^2=1 Considering, x^2/y^2 +y^2/z^2 +z^2/x^2 =(x^4 z^2+y^4 x^2+z^4 y^2)/(〖x^2 y〗^2 z^2 ) And x/y+y/z+z/x=(x^2 z+y^2 x+z^2 y)/xyz=(x^3 〖yz〗^2+y^3 x^2 z+z^3 xy^2)/(〖x^2 y〗^2 z^2 ) As, x^4 z^2+y^4 x^2+z^4 y^2=x^3 〖xz〗^2+y^3 〖yx〗^2+z^3 〖zy〗^2 ≥x^3 〖yz〗^2+y^3 〖xx〗^2+z^3 〖zy〗^2 ≥x^3 〖yz〗^2+y^3 〖zx〗^2+z^3 〖xy〗^2 Therefore, x^2/y^2 +y^2/z^2 +z^2/x^2 ≥x/y+y/z+z/x OK 8 For all x,y,z≥0 and x+y+z=3 1 x=y=z=0 2 x=y=0,z≠0 3 x=0 and y,z≠0 4 x,y,z≠0 From the given0≤ xyz≤1 and 0≤xy+yz+zx≤3 1, and 2 are obvious 3 y+z=3=3/2+3/2, we have y^2 z=(3/2)^2 (3/2)=(3/2)^3=27/8≤4 4 from 0≤xy+yz+zx≤3, 0≤xyz≤1 x+y+z=0, we have 0≤xxy+yyz+zzx+zzx≤3+1=4 So, 〖0≤x〗^2 y+y^2 z+z^2 x+xyz≤4 From 4 cases, it is to be true 9 For all a,b,c,d∈R, a+b+c+d=0 1 a=b=c=d=0 2 a=0,b=-(c+d) 3 a=-(b+c+d) 4 a=-b,c=-d 5 a=b=0,c=-d Case 1,4,and 5 are to be true But case 2,and 3 is to be false Please show to be exercises 10 From {a_n }:3,7,13,21,31,… we see that a_2-a_1=4,a_3-a_2=6,a_4-a_3=8,a_5-a_4=10 (a_3-a_2 )-(a_2-a_1 )=(a_4-a_3 )-(a_3-a_2 ) =(a_5-a_4 )-(a_4-a_3 ) =2 It means that {a_n } be a sequence with quadratic form Is general form Give a_n=bn^2+cn+d will obtain that a_1=b+c+d=3, a_2=4b+2c+d a_3=9b+3c+d=3, a_4=16b+4c+d a_5=25b+5c+d=3, a_6=36b+6c+d From above, we will obtain that b=1,c=1,and d=1 So, a_n=n^2+n+1 a_((n+1) )=(n+1)^2+n+1 =n^2+3n+3 (a_n )(a_((n+1) ) )=n^4+4n^3+7n^2+6n+3 =(n^2+2n+1)^2+(n^2+2n+1)+1 We see that a_k=k^2+k+1,where k=n^2+2n+1,∀n∈N Ok Therefore, we can say that 〖 a〗_i 〖.a〗_((i+1) )∈{a_n } Acknowledgement 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: Thu, 25 Sep 2014 02:53:25 +0000
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