∫ 1/( 1+b+ c)+ 1/( 1+c+a)+ 1/( 1+a+b)≥ 1
∴ Multiply both sides by-1 and add 3 to get:
[ 1- 1/( 1+b+ c)]+[ 1- 1/( 1+c+a)]+[ 1- 1/( 1+a+b)]≤2。
Can be sorted out:
[(b+c)/( 1+b+c)]+[(c+a)/( 1+c+a)]+[(a+b)/( 1+a+b)]≤2
Both sides of this inequality are multiplied by (b+c) (1+b+c)+(c+a)+(a+b) (1+a+b).
free
2[(b+c)( 1+b+c)+(c+a)( 1+c+a)+(a+b)( 1+a+b)]≥
[(b+c)( 1+b+c)+(c+a)( 1+c+a)+(a+b)( 1+a+b)]×[(b+c)/( 1+b+c)+(c+a)/( 1+c+a)+(a+b)/( 1+a+b)]
≥[(b+c)+(c+a)+(a+b)]? . (Cauchy inequality is applied to this step)
∴2[(b+c)( 1+b+c)+(c+a)( 1+c+a)+(a+b)( 1+a+b)]≥4(a+b+c)?
Can be sorted out:
2(a+b+c)+2(a? +b? +c? )+2(ab+bc+ca)≥2(a+b+c)?
∴(a+b+c)+(a? +b? +c? )+(ab+bc+ca)≥a? +b? +c? +2(ab+bc+ca)
∴ Available after finishing:
a+b+c≥ab+bc+ca。