Next 6 problems aren’t mine, but they belong to my Friends - TopicsExpress



          

Next 6 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 If there are someone can write how to prove of some Point, please show for sharing to our friends 1 Give x,y,and z be real numbers that x+y+z=0 write easies form of it 1/(1+2^x+2^(-y) )+1/(1+2^y+2^(-z) )+1/(1+2^z+2^(-x) ) 2 Give x∈(0,π/2), What is the minimum of f(x) f(x)=(sin⁡x )^3/cos⁡x +(cos⁡x )^3/sin⁡x 3 Give p,q,x,and y be positive real numbers And p+q=1,show that (x^2+y^2)/2≥((x+y)/2)^2 and (p+1/p)^2+(q+1/q)^2≥25/2 4 Give x,y be positive real numbers Prove that x/(x^4+y^2 )+y/(y^4+x^2 )≤1/xy 5 Give a,b,c>0,and ab+bc+ca>a+b+c Prove that a+b+c>3 6 Give a,b,c>0 and 1/a+1/b+1/c=√abc Find the minimum value of abc The following writing is my all imagining which It may be correct or incorrect also Solution 1As x+y+z=0, we have 3 cases 1 x=y=z=0 2 x=0,y=-z 3 x+y=-z 4 x=-(y+z) 1 x=y=z=0, 1/(1+2^x+2^(-y) )+1/(1+2^y+2^(-z) )+1/(1+2^z+2^(-x) ) = 1/(1+2^0+2^(-0) )+1/(1+2^0+2^(-0) )+1/(1+2^0+2^(-0) )=1 2 x=0,y=-z, 1/(1+2^x+2^(-y) )+1/(1+2^y+2^(-z) )+1/(1+2^z+2^(-x) ) =1/(1+2^0+2^z )+1/(1+2^(-z)+2^(-z) )+1/(1+2^z+2^(-0) ) =2/(2+2^z )+2^z/(2+2^z )=(2+2^z)/(2+2^z )=1 3 x+y=-z, 1/(1+2^x+2^(-y) )+1/(1+2^y+2^(-z) )+1/(1+2^z+2^(-x) ) =1/(1+2^x+2^(-y) )+1/(1+2^y+2^((x+y)) )+1/(1+2^(-(x+y))+2^(-x) ) =2^y/(1+2^((x+y))+2^y )+1/(1+2^y+2^((x+y)) )+2^((x+y))/(1+2^y+2^((x+y)) ) =(1+2^y+2^((x+y)))/(1+2^((x+y))+2^y )=1 4 x=-(y+z),1/(1+2^x+2^(-y) )+1/(1+2^y+2^(-z) )+1/(1+2^z+2^(-x) ) =1/(1+2^(-(y+z))+2^(-y) )+1/(1+2^y+2^(-z) )+1/(1+2^z+2^((y+z)) ) =2^((y+z))/(2^((y+z))+1+2^z )+2^z/(2^z+2^((y+z))+1)+1/(1+2^z+2^((y+z)) ) =(1+2^z+2^((y+z)))/(1+2^z+2^((y+z)) )=1 2 As x∈(0,π/2), we have (sin⁡x )^3/cos⁡x +(cos⁡x )^3/sin⁡x = ((sin⁡x )^4+(cos⁡x )^4)/cos⁡〖x sin⁡x 〗 Next doing we will write (0,π/2)=(0,π/4]∪[π/4,├ π/2) ┤ And must show that 1 If x∈(0,π/4], then((sin⁡x )^4+(cos⁡x )^4)/cos⁡〖x sin⁡x 〗 ≥1 2 If x∈[π/4,├ π/2) ┤, then((sin⁡x )^4+(cos⁡x )^4)/cos⁡〖x sin⁡x 〗 ≥1 1if x∈(0,π/4], we have 0< sin⁡x≤1/√2, and 1> cos⁡x≥1/√2 0< sin⁡x cos⁡〖x≤1/2〗, and 1/(sin⁡x cos⁡x )≥2 0(cos⁡x )^4≥1/4 1>(sin⁡x )^4+(cos⁡x )^4≥1/2 ((sin⁡x )^4+(cos⁡x )^4)/(sin⁡x cos⁡x )=((sin⁡x )^4+(cos⁡x )^4 )(1/(sin⁡x cos⁡x )) ≥(1/2).2=1 In case 2 is similar, it is left showing as an exercise Therefore, the minimum of f(x) is 1 3 As x^2+y^2≥2xy, it follows 4x^2+〖4y〗^2≥2x^2+4xy+〖2y〗^2 (x^2+y^2)/2≥(x^2+2xy+y^2)/4 (x^2+y^2)/2≥((x+y)/2)^2 2 As p+q=1=1/2+1/2 So, P+ 1/p+q+1/q≥1/2+2+1/2+2 :a/b+b/a≥2,∀a,b>0, 1/p+1/q≥2+2 Hence, (P+ 1/p)^2+(q+1/q)^2≥P+ 1/p+q+1/q ≥(5/2)^2+(5/2)^2=25/2 4 As x,y be positive real numbers By using the fact that a/b+b/a≥2,∀a,b>0 We have, x+1/x+y+1/y≥4 (x^2+1)/x+(y^2+1)/y≥2+2 (x^2+1)/y+(y^2+1)/x≥2+2 (x^4+x^2)/(x^2 y)+(y^4+y^2)/(y^2 x)≥2+2 (x^4+y^2)/(x^2 y)+(y^4+x^2)/(y^2 x)≥2+2 (x^2 y)/(x^4+y^2 )+(y^2 x)/(y^4+x^2 )≤1/2+1/2=1 Therefore, x/(x^4+y^2 )+y/(y^4+x^2 )≤1/xy: OKK 5 As a,b,c>0,and ab+bc+ca>a+b+c Considering (ab+bc+ca)/(a+b+c) =ab/(a+b+c)+bc/(a+b+c)+ca/(a+b+c) ≥ab/3a+bc/3b+ca/3c =b/3+c/3+a/3 =(a+b+c)/3 ≥1: ab+bc+ca>a+b+c Therefore,a+b+c≥3 6 This question is inappropriate, it has to add that a+b+c=k, k>0 As a,b,c>0 and 1/a+1/b+1/c=√abc Considering, 1/a+1/b+1/c=(bc+ca+ab)/abc=√abc abc=(bc+ca+ab)/√abc =bc/√abc+ca/√abc+ab/√abc ≥bc/√bbc+ca/√acc+ab/√aab =bc/(b√c)+ca/(c√a)+ab/(a√b) =c/√c+a/√a+b/√b =√c+√a+√b ≥√(a+b+c) =√k Therefore, the minimum of abc is √k 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, 11 Sep 2014 08:34:25 +0000

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