\(a\left(b-1\right)+b\left(1-c\right)+c\left(1-a\right)\le1\\ \Leftrightarrow-abc+ab+bc+ca-a-b-c+1\le2-abc\\ \Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\le2-abc\)
lại có \(abc\le1\) nên \(2-abc\ge1\)
ta chứng minh \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\le1\)
luôn đúng do \(0\le a;b;c\le1\)
vậy bđt dc cm
tick mik nhaaaaa.mik ms l9 thui