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若a b c d是4个正数,且abcd=1求abc+ab+a+1分之a+bcd+bc+b+1分之b+cda+cd+c+1

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若a b c d是4个正数,且abcd=1求abc+ab+a+1分之a+bcd+bc+b+1分之b+cda+cd+c+1分之c
能减少就减少
若a b c d是4个正数,且abcd=1,求abc+ab+a+1分之a,+bcd+bc+b+1分之b,+cda+cd+c+1分之c,dab+da+d+1分之d
若a b c d是4个正数,且abcd=1求abc+ab+a+1分之a+bcd+bc+b+1分之b+cda+cd+c+1
原式=a/(abc+ab+a+1)+b/(bcd+bc+b+1)+c/(cda+cd+c+1)+d/(dab+da+d+1)
=a/(1/d+1/cd+1/bcd+bcd/bcd)+b/(bcd+bc+b+1)+c/(1/b+1/ab+1/dab+dab/dab)+d/(dab+da+d+1)
=a/(bc/bcd+b/bcd+1/bcd+bcd/bcd)+b/(bcd+bc+b+1)+c/(da/dab+d/dab+1/dab+dab/dab)+d/(dab+da+d+1)
=a/((bcd+bc+b+1)/bcd)+b/(bcd+bc+b+1)+c/((dab+da+d+1)/dab)+d/(dab+da+d+1)
=abcd/(bcd+bc+b+1)+b/(bcd+bc+b+1)+abcd/(dab+da+d+1)+d/(dab+da+d+1)
=(1+b)/(bcd+bc+b+1)+(1+d)/(dab+da+d+1)
=(1+b)/(bcd+bc+b+1)+(1+d)/(1/c+1/bc+bcd/bc+bc/bc)
=(1+b)/(bcd+bc+b+1)+(1+d)/(b/bc+1/bc+bcd/bc+bc/bc)
=(1+b)/(bcd+bc+b+1)+(1+d)/((bcd+bc+b+1)/bc)
=(1+b)/(bcd+bc+b+1)+(bc+bcd)/(bcd+bc+b+1)
=(1+b+bc+bcd)/(bcd+bc+b+1)
=1
∵a,b,c,d都是正数
∴上面过程中所有分母都大于零
∴原式=1