Xét a = b = c = 1 thì thỏa mãn bài ra
Xét a ,b,c khác 1. do a,b,c có vai trò như nhau nên giả sử \(a\le b\le c\)
Áp dụng BĐT cô-si cho 3 số a+b+1,1-a,1-b, ta có :
\(\left(a+b+1\right)\left(1-a\right)\left(1-b\right)\le\left(\frac{a+b+1+1-a+1-b}{3}\right)^3=1\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\le\frac{1}{a+b+1}\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\frac{1-c}{a+b+1}\)
Mà \(\frac{a}{b+c+1}\le\frac{a}{a+b+1};\frac{b}{a+c+1}\le\frac{b}{a+b+1}\)
\(\Rightarrow\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}\le\frac{a}{a+b+1}+\frac{b}{a+b+1}+\frac{c}{a+b+1}\)
do đó : \(\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
\(\le\frac{a+b+c}{a+b+1}+\frac{1-c}{a+b+1}=1\)
dấu " = " xảy ra khi a = b = c = 0
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