12th January 2015 For life’s good start of New Year of all friends Next 13 points aren’t mine but they belong to my Friends which may I praise them as courageous men and my idol 1 give x,y,z>1 and 1/x+1/y+1/z=2, prove that √(x-1)+√(y-1)+√(z-1)≤√(x+y+z) Proof it may have how to do that are easier than this By considering If 1/x+1/y+1/z=2 then x+y+z≥9/2 ( exercise ) And 1/x+1/y≥1/z, 1/y+1/z≥1/x, 1/z+1/x≥1/y Yet I can’t do to succeed, please help me. Proof as 1/x+1/y+1/z=2↔-1/x-1/y-1/z=-2 And x/y+y/z+z/x≥3,y/x+x/z+z/y≥3, we have x/x-1/x+x/y-1/y+x/z-1/z+y/x-1/x+y/y-1/y+ y/z-1/z+z/x-1/x+z/y-1/y+z/z-1/z≥3 And (x-1)/x+(x-1)/y+(x-1)/z+(y-1)/x+(y-1)/y+(y-1)/z+ +(z-1)/x+(z-1)/y+(z-1)/z≥3=9(1/3) And x/(x-1)+y/(x-1)+z/(x-1)+x/(y-1)+y/(y-1)+z/(y-1)+ +x/(z-1)+y/(z-1)+z/(z-1)≥3x9=27 And (x+y+z)/(x-1)+(x+y+z)/(y-1)+(x+y+z)/(z-1)≥9+9+9 And √((x+y+z)/(x-1))+√((x+y+z)/(x-1))+√((x+y+z)/(x-1))≥3+3+3 And 1/√(x-1)+1/√(x-1)+1/√(x-1)≥3/√(x+y+z)+3/√(x+y+z)+3/√(x+y+z) And √(x-1)+√(y-1)+√(z-1)≤ ≤√(x+y+z)/3+√(x+y+z)/3+√(x+y+z)/3=√(x+y+z) That is, √(x-1)+√(y-1)+√(z-1)≤√(x+y+z) 2 give a,b,c>0, prove that (a^(-1) b+b^(-1) c+c^(-1) a)/(a+b+c)≥(a+b+c)/(ab+bc+ca) Proof (c^2+bc)+(a^2+ac)+(b^2+ab) ≥(ab+bc)+(bc+ca)+(ab+ac) And since, 1/a+1/b+1/c=1/a+1/b+1/c So, 1/a (c^2+bc)+1/b (a^2+ac)+1/c(b^2+ab) ≥1/a (ab+bc)+1/a (bc+ca)+1/a(ab+ac) And, a^2/b+ac/b+b^2/c+ab/c+c^2/a+bc/a≥ ≥ab/a+bc/a+bc/b+ca/b+ca/c+ab/c And, a^2/b+ab/b+ac/b+b^2/c+bc/c+ab/c+c^2/a+ca/a+bc/a≥ ≥ab/a+ca/a+bc/a+bc/b+ab/b+ca/b+ca/c+bc/c+ab/c And, b/(a^2+ab+ac)+c/(b^2+bc+ab)+a/(c^2+ca+bc)≥ ≥a/(ab+ab+ac)+b/(ab+bc+ca)+c/(ab+bc+ca) But, b/a(a+b+c) +c/b(a+b+c) +a/c(a+b+c) = =((b/a))/(a+b+c)+((c/b))/(a+b+c)+((a/c))/(a+b+c)= =(a^(-1) b)/(a+b+c)+(b^(-1) c)/(a+b+c)+(c^(-1) a)/(a+b+c) Hence, (a^(-1) b)/(a+b+c)+(b^(-1) c)/(a+b+c)+(c^(-1) a)/(a+b+c)≥(a+b+c)/(ab+bc+ca) 3 give f:N→N, ∀a∈N Defined by f(f(a) )+f(a)=2a+6 Find f Solution As f:N→N, ∀a∈N Thence, f(a) and f(f(a) ) exist And (a,b)∈f↔f(a)=b (b,c)∈f↔f(b)=c So, f(f(a) )=f(b)=c↔(a,c)∈f(f)=fof Seeing, f(f(a) )+f(a)=2a+6 f(b)+b=2a+6 c+b=2a+6 We see that there is only one case b=a+2 and c=a+c Is that, f(a)=b=a+2 f(f(a))= f(b)=f(a+2)=a+4 =c And, c+b=a+4+a+2=2a+6 We find that if b≠a+2 will make c≠a+4 And result in c+b≠2a+6 That is, f(f(a) )+f(a)≠2a+6 Therefore, f(x)=x+2,∀x∈N 4 give a,b,c>0 prove that (1/a+1/b+1/c)(1/(a+1)+1/(b+1)+1/(c+1))≥9/(1+abc) Proof As ab+a+1/c+1/bc+bc+b+ +1/a+1/ac+ac+c+1/b+1/ab≥12 So, ab+a+ab/abc+a/abc+bc+b+ +bc/abc+b/abc+ac+c+ac/abc+c/abc≥12 And, ab+a+(ab+a)/abc+bc+b+(bc+b)/abc+ +ca+c+(ac+c)/abc≥12 And, 1/(ab+a)+abc/(ab+a)+1/(bc+b)+abc/(bc+b)+ +1/(ac+c)+abc/(ac+c)≥1/2+1/2+1/2+1/2+1/2+1/2 And, ((1+abc)/(ab+a))+((1+abc)/(bc+b))+((1+abc)/(ac+c))≥3=1+1+1 And, ((1+abc)/(ab+a))((1+abc)/(bc+b))((1+abc)/(ac+c))≥1 And (1+abc)^3/(abc)(b+1)(c+1)(a+1) ≥1 And, (1/abc)(1/(b+1)(c+1)(a+1) )≥1/(1+abc)^3 And 3(∛(1/abc))3(∛(1/(b+1)(c+1)(a+1) ))≥9/(1+abc) Hence, (1/a+1/b+1/c)(1/(a+1)+1/(b+1)+1/(c+1))≥9/(1+abc) 5 ∀x,y,z>0 show that 2/(x+y)+2/(y+z)+2/(z+x)≥9/(x+y+z) Proof As 1+ y/x+x/y+1+x/z+y/z+y/x+z/x+1+ +z/y+1+y/z+z/x+1+z/y+x/y+1+x/z≥18 And, (x+y)/x+(x+y)/y+(x+y)/z+(y+z)/x+(y+z)/y+(y+z)/z+ +(z+x)/x+(z+x)/y+(z+x)/z≥2+⋯+2 And, x/(x+y)+y/(x+y)+z/(x+y)+x/(y+z)+y/(y+z)+z/(y+z)+ +x/(z+x)+y/(z+x)+z/(z+x)≥1/2+..+1/2 And, (x+y+z)/(x+y)+(x+y+z)/(y+z)+(x+y+z)/(z+x)≥3/2+3/2+3/2=9/2 Hence, 2/(x+y)+2/(y+z)+2/(z+x)≥9/(x+y+z) 6 give a,b,c be the lengths of the sides of a triangle Prove that a^3+3abc+b^3≥c^3 Proof As a+b>c, we have (a+b)^3=a^3+3a^2 b+3ab^2+b^3 =a^3+3ab(a+b)+b^3 ≥a^3+3abc+b^3 =(a+b)(a^2-ab+b^2 )+3abc ≥c(a^2-ab+b^2 )+3abc =a^2 c-abc+b^2 c+3abc =ac(a+b)+bc(b+a) ≥ac^2+bc^2 =c^2 (a+b) ≥c^2 c=c^3 yes 7 find every Prime numbers p such that p+10 and p+20 are also Prime simultaneous. Solution by Division Algorithm, we have ∀n∈z, n=3k,3k+1 and 3k+2,∃k∈z Hence, 1 if p∈3k, then p=3x1=3 And, 3+10,3+20 are alo Prime simultaneous. But if p∈3k+1,3k+2, then 3k+1+20,and,3k+2+10 aren^ t Peime. Therefore, there is one Prime to be 3 only. 8 find the next sequence 5,7,11,13,17,19,23,25,29,31,35,… Solution Seeing, when n∈odd 5,11,17,23,29,35,… Give a_n=nx+y 5=x+y 11=3x+y Hence, x=3,y=2 a_n=3n+2 And when n∈even 7,13,19,25,31,37,43,.. Give a_n=nx+y 7=x+y 13=3x+y Hence, x=3,y=1 a_n=3n+1 Therefore, a_n={█(3n+2,n∈odd@ 3n+1,n∈even)┤ 9 give a,b,c>0 and a+b+c=6abc Prove that 1/(a^2+b^2+1)+ 1/(b^2+c^2+1)+1/(c^2+a^2+1)≤3/2 Proof a+b+c=6abc ↔1/6bc+1/6ac+1/6ab=1 ↔6bc+6ac+6ab≥9 ↔bc+ac+ab≥3/2 ↔a^2+b^2+c^2≥1/2+1/2+1/2 ↔a^2+b^2+1+b^2+c^2+1+c^2+a^2+1≥6 ↔1/(a^2+b^2+1)+ 1/(b^2+c^2+1)+1/(c^2+a^2+1)≤1/2+1/2+1/2=3/2 Hence, 1/(a^2+b^2+1)+ 1/(b^2+c^2+1)+1/(c^2+a^2+1)≤3/2 ok 10 give a,b,c be the lengths of the sides of a triangle Prove that 1/(a+b), 1/(b+c), 1/(c+a) be same to be above. Proof we must show that 1/(a+b)+ 1/(b+c)> 1/(c+a), 1/(a+b)+ 1/(c+a)> 1/(b+c), and 1/(b+c)+ 1/(c+a)> 1/(a+b) We will show 1/(a+b)+ 1/(b+c)> 1/(c+a) only Remaining cases are left the proofs as exercises. As a^2+c^2+ab+ac+bc>b^2:b(a+c)>b^2 So, a^2+c^2+2ab+2ac+2bc≥b^2+ab+bc+ca And, (a^2+c^2+2ab+2ac+2bc)/(b^2+ab+bc+ca)>1 And, ((c+a)(b+c)+(c+a)((a+b))/((a+b)(b+c))>1 And, (c+a)(b+c)/((a+b)(b+c))+((c+a)((a+b))/((a+b)(b+c))>1 And, ((c+a))/((a+b) )+((c+a))/((b+c))>1 Therefore, 1/(a+b)+ 1/(b+c)> 1/(c+a) yes 11 give a,b,c,d≥0, show that √(a+b+c)+√(b+c+d)+√(c+d+a)+ +√(d+a+b)≥3√(a+b+c+d) Proof 1 when a=b=c=d=0 We see that it is to be true. 2 when a=b=c=0,d≠0 We see that it is to be true. 3 when a=b=0,c,d≠0, we have √c+√(c+d)+√(c+d)+√d≥3√(c+d) Is to be true, as √c+√d≥√(c+d) 4 when a=0,b,c,d≠0, we have As 1/2=2/4=(2(b+c+d))/(4(b+c+d))= = (b+c)/(4(b+c+d))+(c+d)/(4(b+c+d))+(d+b)/(4(b+c+d))≥1/9+1/9+1/9 And, √(b+c)/(2√(b+c+d))+√(c+d)/(2√(b+c+d))+√(d+b)/(2√(b+c+d))≥1/3+1/3+1/3 And, √(b+c)+√(c+d)+√(d+b)≥2√(b+c+d) So, √(b+c)+√(b+c+d)+√(c+d)+ +√(d+b)≥3√(b+c+d) ok 5 when a,b,c,d≠0 is an exercise. 12 show that 3⁄((n^3+(n+1)^3+(n+2)^3 ) ) Solution from Division Algorithm, we have n=3k,or n=3k+1,or n=3k+2,∃∈z 1if n=3k→n^3+(n+1)^3+(n+2)^3 =(3k)^3+(3k+1)^3+(3k+2)^3 =81k^3+81k^2+45k+9 =3(27k^3+27k^2+15k+3) 2 if n=3k+1→n^3+(n+1)^3+(n+2)^3 =(3k+1)^3+(3k+2)^3+(3k+3)^3 =54k^3+81k^2+45k+9+27(k+1)^3 =3(18k^3+27k^2+15k+3+9(k+1)^3) 3 if n=3k+2→n^3+(n+1)^3+(n+2)^3 =(3k+2)^3+(3k+3)^3+(3k+4)^3 =54k^3+162k^2+180k+72+27(k+1)^3 =3(18k^3+54k^2+60k+24+9(k+1)^3) From 3 cases, we will obtain that 3⁄((n^3+(n+1)^3+(n+2)^3 ) ) ok 13 give a,b,c≥0,and a+b+c=12 Prove that (a^2+a+1)(b^2+b+1)(c^2+c+1)≤(21)^3 Proof As a+b+c≥3(∛abc) So, 0≤3(∛abc)≤12 0≤(∛abc)≤4 0≤abc≤64 0≤a^2 b^2 c^2≤(64)^2=16x16x16 And, a+b+c=4+4+4→0≤abc≤4x4x4 And, 0≤(a^2+a)(b^2+b)(c^2+c)≤20x20x20 And, 157≤(a^2+a+1)(b^2+b+1)(c^2+c+1) ≤21x21x21=(21)^3 Therefore, (a^2+a+1)(b^2+b+1)(c^2+c+1)≤(21)^3 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: Mon, 12 Jan 2015 07:32:42 +0000
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