Giải:
Vì \(0\leq a,b,c\leq 1\Rightarrow ab,ac,ab\geq abc\)
Do đó mà \(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq \frac{a+b+c}{abc+1}\)
Giờ chỉ cần chỉ ra \(\frac{a+b+c}{abc+1}\leq 2\). Thật vậy:
Do \(0\leq b,c\leq 1\Rightarrow (b-1)(c-1)\geq 0\Leftrightarrow bc+1\geq b+c\Rightarrow bc+a+1\geq a+b+c\)
Suy ra \( \frac{a+b+c}{abc+1}\leq \frac{bc+a+1}{abc+1}=\frac{bc+a-2abc-1}{abc+1}+2=\frac{(bc-1)(1-a)-abc}{abc+1}+2\)
Ta có \(\left\{\begin{matrix}bc\le1\\a\le1\\abc\ge0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}\left(bc-1\right)\left(1-a\right)\le1\\-abc\le0\end{matrix}\right.\) \(\Rightarrow \frac{(bc-1)(1-a)-abc}{abc+1}+2\leq 2\Rightarrow \frac{a+b+c}{abc+1}\leq 2\)
Chứng minh hoàn tất
Dấu bằng xảy ra khi \((a,b,c)=(0,1,1)\) và hoán vị.
