25th December 2014 The Christmas Day and New Year Day for - TopicsExpress

25th December 2014 The Christmas Day and New Year Day for All’s day Next 13 points aren’t mine but I believe that all maker Problems and solver problems are supported each others 1 this point almost can do it, but I want to use ln⁡x Find the answer in R of (x√x)^x=(x)^(x√x) Solution from(x√x)^x=(x)^(x√x),we have ln⁡〖(x√x)^x 〗=ln⁡〖(x)^(x√x) 〗 x(ln⁡〖x√x〗 )=x√x ln⁡x x(ln⁡x+ln⁡√x )=x√x ln⁡x x ln⁡x+1/2 x ln⁡x=x√x ln⁡x x ln⁡x+1/2 ln⁡x-x√x ln⁡x=0 ln⁡x (x+1/2 x-x√x)=0 Hence, ln⁡x=0 or x+1/2-x√x=0 ln⁡x=0↔x=1 and x+1/2-x√x=0 →2x+x-2 x√x=0 →x(3-2√x)=0 →x=0,9/4 From rechecking solution, its answer is {1,9/4 } 2 give x,y,z>0, Prove that x^2/y^2 +y^2/z^2 +z^2/x^2 +8(xy+yz+zx)/(x^2+y^2+z^2 )≥11 Proof As 11- 8(xy+yz+zx)/(x^2+y^2+z^2 )≥3 and x^2/y^2 +y^2/z^2 +z^2/x^2 ≥3 But, 0< (xy+yz+zx)/(x^2+y^2+z^2 )≤1,we have -1≤ (-(xy+yz+zx))/(x^2+y^2+z^2 )0, prove that a/b+b/c+c/a≥(a+b)/(b+c)+(b+c)/(c+a)+(c+a)/(a+b) Proof As 2b+2c+2d+2a= (b+c)+(a+b)+(c+d)+(d+a) So, 1/2b+1/2c+1/2d+1/2a≥1/(b+c)+1/(a+b)+1/(c+d)+1/(d+a) And, 1/b+1/c+1/d+1/a≥2/(b+c)+2/(a+b)+2/(c+d)+2/(d+a) =(1/(b+c)+1/(a+b))+(1/(b+c)+1/(c+d))+(1/(c+d)+1/(d+a))+(1/(d+a)+1/(a+b)) So, a/b+b/c+c/d+d/a≥ ≥(a/(b+c)+a/(a+b))+(b/(b+c)+b/(c+d))+(c/(c+d)+c/(d+a))+(d/(d+a)+d/(a+b)) =(a/(b+c)+b/(b+c))+(b/(c+d)+c/(c+d))+(c/(d+a)+d/(d+a))+(d/(a+)+a/(a+b)) =((a+b)/(b+c))+((b+c)/(c+d))+((c+d)/(d+a))+((d+a)/(a+)): OK Remark: a/b+b/c+c/d+d/a:A (a+b)/(b+c)+(b+c)/(c+a)+(c+a)/(a+b):B A≥B↔1/b+1/c+1/d+1/a≥ ( important) (1/(b+c)+1/(a+b))+(1/(b+c)+1/(c+d))+(1/(c+d)+1/(d+a))+(1/(d+a)+1/(a+b)) 9 give a,b,c>0,prove that a^4/(a^2+ab+b^2 )+b^4/(b^2+bc+c^2 )+c^4/(c^2+ca+a^2 )≥(a^3+b^3+c^3)/(a+b+c) Proof as 2a^2+〖2b〗^2+〖2c〗^2+ab+bc+cd≥ a^2+b^2+c^2+2ab+2bc+2cd And a^2+ab+b^2≥a^2+ab+ac Hence, (a^2+ab+b^2)/a^4 ≥(a+b+c)/a^3 , similar (b^2+bc+c^2)/b^4 ≥(a+b+c)/b^3 , and (c^2+ca+a^2)/c^4 ≥(a+b+c)/c^3 So, (a^2+ab+b^2)/a^4 +(b^2+bc+c^2)/b^4 (c^2+ca+a^2)/c^4 ≥ (a+b+c)/a^3 +(a+b+c)/b^3 +(a+b+c)/c^3 And make, a^4/(a^2+ab+b^2 )+b^4/(b^2+bc+c^2 )+c^4/(c^2+ca+a^2 )≥ a^3/(a+b+c)+b^3/(a+b+c)+c^3/(a+b+c)=(a^3+b^3+c^3)/(a+b+c):OK 10 Prove that (a^3+b^3+c^3)/(a+b+c)+(a^3+b^3+d^3)/(a+b+d)+(a^3+c^3+d^3)/(a+c+d)+ + (b^3+c^3+d^3)/(b+c+d)≥a^2+b^2+c^2+d^2 Proof As 2(a^3+b^3+c^3+d^3)≥ a^2 b+b^2 d+d^2 c+c^2 a+a^2 c+c^2 d+d^2 b+b^2 a And, b^3+c^3+a^3+d^3+a^3+d^3+b^3+c^3≥ a^2 b+b^2 d+d^2 c+c^2 a+a^2 c+c^2 d+d^2 b+b^2 a And (a^3 〖+b〗^3+c^3 )+(a^3 〖+b〗^3+d^3 )+ +(a^3 〖+c〗^3+d^3 )+(b^3 〖+c〗^3+d^3 )≥a^3+a^2 b+a^2 c+ +b^2 a+b^3+b^2 d+c^2 a+c^3+c^2 d+d^2 b+d^2 c+d^3 =a^2 (a+b+c)+b^2 (a+b+d)+c^2 (a+c+d) +d^2 (b+c+d) Next, we can write a^3 〖+b〗^3+c^3≥a^2 (a+b+c), or a^3 〖+b〗^3+d^3≥b^2 (a+b+d), or a^3 〖+c〗^3+d^3≥c^2 (a+c+d),or b^3 〖+c〗^3+d^3≥d^2 (b+c+d) Hence, (a^3 〖+b〗^3+c^3)/(a+b+c)≥a^2,or (a^3 〖+b〗^3+d^3)/(a+b+d)≥b^2 (a^3 〖+c〗^3+d^3)/(a+c+d)≥c^2,or (b^3 〖+c〗^3+d^3)/(b+c+d)≥d^2 Therefore, (a^3+b^3+c^3)/(a+b+c)+(a^3+b^3+d^3)/(a+b+d)+ (a^3+c^3+d^3)/(a+c+d)+ + (b^3+c^3+d^3)/(b+c+d)≥a^2+b^2+c^2+d^2 11 give a,b,c>0, prove that√(a^2+8bc) +√(b^2+8ca)+√(c^2+8ab)≤3(a+b+c) Proof As a^2+8bc+b^2+8ca+c^2+8ab≤ 3(a+b+c)^2 So, we write, a^2+8bc≤(a+b+c)^2,or b^2+8ca≤(a+b+c)^2, or, c^2+8ab≤(a+b+c)^2 And, √(a^2+8bc)≤a+b+c, or √(b^2+8ca)≤a+b+c, or √(c^2+8ab)≤a+b+c Therefore, √(a^2+8bc)+ +√(b^2+8ca)+√(c^2+8ab)≤3(a+b+c) 12find the solution of x+2√(7-x)=2√(x-1)+√(-x^2+8x-7)+1 Solution give m=√(7-x),n=√(x-1), we get x-1+2√(7-x)=2√(x-1)+√((x-1)(7-x) ) n^2+2m=2n+mn n^2-2n=mn-2m n(n-2)=m(n-2) Hence, n=m,where n≠2 √(7-x)=√(x-1),√(x-1)≠2 7-x= x-1,x-1≠4 x=4,x≠5 Recheck its solution 4+2√(7-4)=2√(4-1)+√(-4^2+8x4-7)+1 4+2√3=2√3+4 13 give f(x)+f(f(x) )=2x+6,∀x∈N Find f(2015) Solution As f is linear equation So, f(x)=ax+b, we get f(f(x))=a(ax+b)+b =a^2 x+ab+b Seeing, f(x)+f(f(x) )=2x+6 ax+b+a^2 x+ab+b=2x+6 From solving, we get a=1,and b=2 Hence, f(x)=x+2 Recheck: f(x)+f(f(x) )= x+2+x+4=2x+6 And f(2015)=2015+2=2017 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