Question

cho các số thực dương a,b,c thỏa mãn \(a+b+c=6\). Chứng minh rằng: 

\(\frac{b+c+5}{1+a}+\frac{c+a+4}{2+b}+\frac{a+b+3}{3+c}\ge6\)

Answer 1

Ta có \(\frac{b+c+6}{1+a}=\frac{11-a}{1+a}=-1+\frac{12}{1+a}\)

\(\frac{c+a+4}{2+b}=-1+\frac{12}{2+b}\)

\(\frac{a+b+3}{3+c}=-1+\frac{12}{3+c}\)

Mà \(\frac{1}{1+a}+\frac{1}{2+b}+\frac{1}{3+c}\ge\)

\(\frac{3^2}{1+2+3+a+b+c}=\frac{3}{4}\)

Từ đó => VT \(\ge\)-3 + \(12\frac{3}{4}\)= 6

Answer 2

Đặt x=a+1; y=b+2; z=3+c (x;y;z>0)

\(VT=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

\(=\frac{y}{x}+\frac{x}{y}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}\)

\(\ge2\sqrt{\frac{y}{x}\cdot\frac{x}{y}}+2\sqrt{\frac{z}{x}\cdot\frac{x}{z}}+2\sqrt{\frac{y}{z}\cdot\frac{z}{y}}=6\)

Dấu "=" xảy ra <=> a=3; b=2; c=1