Chứng minh 1/a(a+1)+1/b(b+1)+1/c(c+1)>=3/(căn bậc 3(abc)(1+căn bậc 3(abc)))

Lời giải:

Áp dụng hệ quả của BĐT AM-GM:

\(\text{VT}^2=\left[\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\right]^2\geq 3\left(\frac{1}{ab(a+1)(b+1)}+\frac{1}{bc(b+1)(c+1)}+\frac{1}{ca(a+1)(c+1)}\right)\)

\(\Leftrightarrow \text{VT}^2\geq 3.\frac{a^2+b^2+c^2+a+b+c}{abc(a+1)(b+1)(c+1)}\geq 3.\frac{a+b+c+ab+bc+ac}{abc(a+1)(b+1)(c+1)}\)

\(\Leftrightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(a+1)(b+1)(c+1)}\) \((1)\)

Ta sẽ cm \((a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3\). Thật vậy:

Áp dụng BĐT AM-GM:

\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\)

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}\)

Cộng theo vế: \(\Rightarrow 3\geq \frac{3(\sqrt[3]{abc}+1)}{\sqrt[3]{(a+1)(b+1)(c+1)}}\)

\(\Rightarrow (a+1)(b+1)(c+1)\geq (\sqrt[3]{abc}+1)^3\) (2)

Từ \((1),(2)\Rightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(1+\sqrt[3]{abc})^3}=\frac{9}{\sqrt[3]{a^2b^2c^2}(1+\sqrt[3]{abc})^2}=\text{VP}^2\)

\(\Leftrightarrow \text{VT}\geq \text{VP}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)