Question

Cho a,b,c>0 có a+b+c=1. CMR:

\(\frac{19b^3-a^3}{ba+5b^2}+\frac{19c^3-b^3}{bc+5c^2}+\frac{19a^3-c^3}{ac+5a^2}\le3\)

Answer 1

Cần chứng minh: \(\frac{19b^3-a^3}{ab+5b^2}\le4b-a\)

Thật vậy: \(\frac{19b^3-a^3}{ab+5b^2}\le4b-a\Leftrightarrow\left(4b-a\right)\left(ab+5b^2\right)-19b^3+a^3\ge0\)

\(\Leftrightarrow4ab^2+20b^3-a^2b-5ab^2-19b^3+a^3\ge0\)

\(\Leftrightarrow\left(a^3+b^3\right)-ab\left(a+b\right)\ge0\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)(đúng)

"=" khi a=b

Tương tự: \(\frac{19c^3-b^3}{bc+5c^2}\le4c-b;\frac{19a^3-c^3}{ac+5a^2}\le4a-c\)

Cộng theo vế: 

\(\frac{19b^3-a^3}{ab+5b^2}+\frac{19c^3-b^3}{bc+5c^2}+\frac{19a^3-c^3}{ac+5a^2}\le4b-a+4c-b+4a-c=3\left(a+b+c\right)=3\)

Dấu "=" xảy ra khi a=b=c=1/3