Question

Cho a,,b,c >0

1/a+b+1/b+c+1/c+a>=2(1/2a+b+c+1/a+2b+c+1/a+b+2c)

Answer 1

Với các số dương, áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

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

Cộng vế với vế:

\(2\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge4\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)

\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\ge2\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)