Question

[Hung Trinh Ngoc]

Cho a,b,c>0.

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

Answer 1

[Thắng Nguyễn]

Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có:

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

Tương tự cho 2 BĐT còn lại ta cũng có:

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

Cộng theo vế 3 BĐT trên ta có: 

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

\(\ge\frac{1}{a+b+2c}+\frac{1}{2a+b+c}+\frac{1}{a+2b+c}=VP\)

Answer 2

[Hung Trinh Ngoc]

thanks