Lời giải:
Đặt biểu thức vế trái là $A$
Ta có:
\(A+3=\frac{b+c+5}{a+1}+1+\frac{a+c+4}{b+2}+1+\frac{a+b+3}{c+3}+1\)
\(=\frac{a+b+c+6}{a+1}+\frac{a+b+c+6}{b+2}+\frac{a+b+c+6}{c+3}\)
\(=(a+b+c+6)\left(\frac{1}{a+1}+\frac{1}{b+2}+\frac{1}{c+3}\right)\)
Áp dụng BĐT Cauchy-Schwarz hay (Svac-sơ) ta có:
\(\frac{1}{a+1}+\frac{1}{b+2}+\frac{1}{c+3}\geq \frac{9}{a+1+b+2+c+3}=\frac{9}{a+b+c+6}\)
\(\Rightarrow A+3\geq (a+b+c+6).\frac{9}{a+b+c+6}=9\Rightarrow A\geq 6\) (đpcm)