Question

Cho a,b,c \(\in R^+\) và a.b.c=1. Chứng minh rằng:

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

Answer 1

Lời giải:

\(\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq \frac{3}{4}\)

\(\Leftrightarrow \frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}\geq \frac{3}{4}\)

\(\Leftrightarrow 4[a(c+1)+b(a+1)+c(b+1)]\geq 3(a+1)(b+1)(c+1)\)

\(\Leftrightarrow 4(ab+bc+ac+a+b+c)\geq 3[(ab+bc+ac)+(a+b+c)+abc+1]\)

\(\Leftrightarrow ab+bc+ac+a+b+c\geq 3(abc+1)=6\)

Điều này luôn đúng do theo BĐT AM-GM thì \(ab+bc+ac+a+b+c\geq 6\sqrt[6]{(abc)^3}=6\)

Ta có đpcm. Dấu "=" xảy ra khi $a=b=c=1$