Question

[Cố Gắng Hơn Nữa]

a;b;c>0 / abc=1. CMR:

\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)

Answer 1

[Akai Haruma]

Lời giải:

Ta có:

\(\text{VT}=\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\)

\(=\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}=\frac{ab+bc+ac+a+b+c}{abc+(ab+bc+ac)+(a+b+c)+1}\)

\(=\frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\)

Ta cần chứng minh \(\text{VT}\geq \frac{3}{4}\)

\(\Leftrightarrow \frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\geq \frac{3}{4}\)

\(\Leftrightarrow 4(ab+bc+ac+a+b+c)\geq 3(ab+bc+ac+a+b+c)+6\)

\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\)

\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\sqrt[6]{ab.bc.ac.a.b.c}\)

(Đúng theo BĐT Cô-si)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c=1\)

Answer 2

[Cố Gắng Hơn Nữa]

em cảm ơn nhiều nha