Question

[Lâm Ngọc]

cho a,b,c>0 thỏa mãn\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=2\)2 .CMR ab+bc+ac\(\ge\)12

Answer 1

[Kiệt Nguyễn]

Ta có: \(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=2\Leftrightarrow\left(1-\frac{a}{a+1}\right)+\left(1-\frac{b}{b+1}\right)+\left(1-\frac{c}{c+1}\right)=1\)\(\Leftrightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1\)\(\Leftrightarrow\left(b+1\right)\left(c+1\right)+\left(c+1\right)\left(a+1\right)+\left(a+1\right)\left(b+1\right)=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)\(\Leftrightarrow a+b+c+2=abc\ge3\sqrt[3]{abc}+2\)(Bất đẳng thức Cô - si)

Đặt \(t=\sqrt[3]{abc}\)thì \(t^3\ge3t^3+2\Leftrightarrow\left(t-2\right)\left(t+1\right)^2\ge0\Leftrightarrow t\ge2\)(Do \(\left(t+1\right)^2>0\forall t>0\))

\(\Rightarrow abc\ge8\)

\(\Rightarrow ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\ge12\)(đpcm)

Đẳng thức xảy ra khi a = b = c = 2