+)a=b=c=0 (thỏa mãn)
+) a+b+c \(\ne\) 0
Ta có: \(a\le1;b-1\le0;c-1\le0;bc\ge0\)
\(\Leftrightarrow a\le1;\left(b-1\right)\left(c-1\right)\ge0;bc\ge0\)
\(\Leftrightarrow a\le1;\left(bc+1\right)-\left(b+c\right)\ge0;bc\ge0\)
\(\Leftrightarrow a\le1;bc+1\ge b+c;bc\ge0\)
\(\Leftrightarrow a+b+c\le bc+bc+1+1=2bc+2=2\left(bc+1\right)\)
\(màa+b+c<0\left(a+b+c\ne0\right)\)
\(\Rightarrow\frac{1}{bc+1}\le\frac{2}{a+b+c}\Rightarrow\frac{a}{bc+1}\le\frac{2a}{a+b+c}\)
CMTT với \(\frac{b}{ac+1}\le\frac{2b}{a+b+c};\frac{c}{ab+1}\le\frac{2c}{a+b+c}\) ,sử dụng phương pháp cộng từng vế=>đpcm