Question

[Linh Nguyễn]

cho 3 số dương a,b,c.chung minh:\(\frac{b}{a+3b}+\frac{c}{b+3c}+\frac{a}{c+3a}\le\frac{3}{4}\)

Answer 1

[Aeris]

Với a, b, c là các số dương, áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{b}{a+3b}+\frac{c}{b+3c}+\frac{a}{c+3a}=\frac{a+b+c}{a+3b+b+3c+c+3a}=\frac{a+b+c}{4a+4b+4c}=\frac{a+b+c}{4\left(a+b+c\right)}=\frac{1}{4}\)

\(\Rightarrow\frac{b}{a+3b}+\frac{c}{b+3c}+\frac{a}{c+3a}\le\frac{3}{4}\)

Answer 2

[Lê Anh Tú]

Ta có: \(\frac{b}{a+3b}+\frac{c}{b+3c}+\frac{a}{c+3a}\le\frac{3}{4}\)(*)

\(\Leftrightarrow3ba+3b+\frac{3c}{b+3c}+\frac{3a}{c+3a}\le\frac{9}{4}\)

\(\Leftrightarrow1-3ba+3b+1-\frac{3c}{b+3c}+1-\frac{3a}{c+3a}\ge\frac{3}{4}\)

\(\Leftrightarrow\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)

Áp dụng BĐT Cauchy - swarch có

\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}=\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3ab+3bc+3ca}\)

Ta sẽ chứng minh : \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3ab+3bc+3ca}\ge\frac{3}{4}\left(1\right)\)

Từ (1) \(\Leftrightarrow4\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2+3ab+3bc+3ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)(đúng)

Vậy (*) đúng