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Question Number 196629 by universe last updated on 28/Aug/23
if xyz=1, prove  ((x/(x−1)))^2 +((y/(y−1)))^2 +((z/(z−1)))^2 ≥1.
ifxyz=1,prove(xx1)2+(yy1)2+(zz1)21.
Commented by universe last updated on 28/Aug/23
  solution to question #196519
You can't use 'macro parameter character #' in math mode
Answered by universe last updated on 28/Aug/23
       let   a= (x/(1−x))  , b= (y/(1−y))  ,  c  =  (z/(1−z))     abc  = ((xyz)/((1−x)(1−y)(1−z))) = (1/((1−x)1−y)(1−z)))     a+1 = (1/(1−x)) , b+1 = (1/(1−y))  , c+1 = (1/(1−z))  (a+1)(b+1)(c+1) = (1/((1−x)(1−y)(1−z)))  (a+1)(b+1)(c+1) = abc  abc = abc+ab+bc+ca+a+b+c+1  ab+bc+ca+a+b+c+1 = 0  2(ab+bc+ca)+2(a+b+c) + 2 = 0  a^2 +b^2 +c^2 +2(ab+bc+ca)+2(a+b+c)+2 = a^2 +b^2 +c^2   (a+b+c)^2 +2(a+b+c)+2 = a^2 +b^2 +c^2     (a+b+c+1)^2 +1 = a^2 +b^2 +c^2       now  (a+b+c+1)^2 +1 ≥1      so  a^2 +b^2 +c^2 ≥1      ( (x/(1−x)))^2 +((y/(1−y)))^2 +((z/(1−z)))^2 ≥ 1
leta=x1x,b=y1y,c=z1zabc=xyz(1x)(1y)(1z)=1(1x)1y)(1z)a+1=11x,b+1=11y,c+1=11z(a+1)(b+1)(c+1)=1(1x)(1y)(1z)(a+1)(b+1)(c+1)=abcabc=abc+ab+bc+ca+a+b+c+1ab+bc+ca+a+b+c+1=02(ab+bc+ca)+2(a+b+c)+2=0a2+b2+c2+2(ab+bc+ca)+2(a+b+c)+2=a2+b2+c2(a+b+c)2+2(a+b+c)+2=a2+b2+c2(a+b+c+1)2+1=a2+b2+c2now(a+b+c+1)2+11soa2+b2+c21(x1x)2+(y1y)2+(z1z)21
Commented by mr W last updated on 28/Aug/23
great! thanks!
great!thanks!

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