Question

cho a,b,c>0 thỏa mãn \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge2\). Cmr: abc ≤ \(\frac{1}{8}\)

Answer 1

Từ đề bài suy ra \(\frac{1}{a+1}\ge\left(1-\frac{1}{b+1}\right)+\left(1-\frac{1}{c+1}\right)=\frac{b}{b+1}+\frac{c}{c+1}\ge2\sqrt{\frac{bc}{\left(b+1\right)\left(c+1\right)}}\)

Tương tự với hai bđt kia rồi nhân theo vế suy ra

\(\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{8abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

Do a, b, c>0 nên (a+1)(b+1)(c+1) > 0 suy ra:

\(1\ge8abc\Leftrightarrow abc\le\frac{1}{8}\left(đpcm\right)\)

Đẳng thức xảy ra khi a = b = c = 1/2