đặt \(A=\frac{b+c+5}{a+1}+\frac{c+a+4}{b+2}+\frac{a+b+3}{c+3}\)
\(=\frac{12-\left(a+1\right)}{a+1}+\frac{12-\left(b+2\right)}{b+2}+\frac{12-\left(c+3\right)}{c+3}\)
\(=\frac{12}{a+1}+\frac{12}{b+2}+\frac{12}{c+3}-3\ge\frac{108}{a+b+c+1+2+3}-3=\frac{108}{12}-3=6\)(Q.E.D)
dấu = xảy ra khi a+1=b+2=c+3<=>a=3;b=2;c=1