Question

Cho a, b, c > 0. Chứng minh rằng: \(\dfrac{19b^3-a^3}{ab+5b^2}+\dfrac{19c^3-b^3}{cb+5c^2}+\dfrac{19a^3-c^3}{ac+5a^2}\le3\left(a+b+c\right)\)

Answer 1

Chuẩn hóa: a+b+c=3k

\(\Rightarrow\)\(\dfrac{a}{k}+\dfrac{b}{k}+\dfrac{c}{k}=3\)

Đặt (\(\dfrac{a}{k};\dfrac{b}{k};\dfrac{c}{k}\))\(\Rightarrow\left(x;y;z\right)\);x+y+z=3

ĐPCM\(\Leftrightarrow\)\(\sum\dfrac{19y^3-x^3}{xy+5y^2}\le3\left(x+y+z\right)\)

Ta CM BĐT:

\(\dfrac{19y^3-x^3}{xy+5y^2}\le4y-x\Leftrightarrow-\dfrac{\left(y-x\right)^2\left(x+y\right)}{xy+5y^2}\le0\)(đúng)

CMTT\(\Rightarrow\)ĐPCM