Lời giải:
Áp dụng BĐT AM-GM cho các số dương ta có:
\(\frac{a^3}{(a+1)(b+1)}+\frac{a+1}{8}+\frac{b+1}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(b+1)(c+1)}+\frac{b+1}{8}+\frac{c+1}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(c+1)(a+1)}+\frac{c+1}{8}+\frac{a+1}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng theo vế:
\(\Rightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{a+b+c+3}{4}\geq \frac{3}{4}(a+b+c)\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{3}{2}\geq \frac{9}{4}\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
