Question

[tep.]

Cho các số thực dương a,b,c thỏa mãn a+b+c=3. Chứng minh rằng ∑\(\dfrac{1}{a+3b}\)≥ ∑\(\dfrac{1}{a+3}\)

Answer 1

[Nguyễn Việt Lâm]

Ta có:

\(\dfrac{1}{a+3b}+\dfrac{1}{c+3}\ge\dfrac{4}{a+3b+c+3}=\dfrac{4}{2b+6}=\dfrac{2}{b+3}\)

Tương tự: 

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

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

Cộng vế:

\(\sum\dfrac{1}{a+3b}+\sum\dfrac{1}{a+3}\ge\sum\dfrac{2}{a+3}\)

\(\Rightarrow\sum\dfrac{1}{a+3b}\ge\sum\dfrac{1}{a+3}\) (đpcm)