Ta có: \(a^2+b^2-ab\ge ab\)
\(\Leftrightarrow\left(a+b\right)\left(a^2+b^2-ab\right)\ge ab\left(a+b\right)\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)
\(\Rightarrow a^3+20b^3\ge19b^3+ab\left(a+b\right)\Leftrightarrow20b^3-ab\left(a+b\right)\)\(\ge19b^3-a^3\)
\(\Leftrightarrow b\left(20b^2-ab-a^2\right)\ge19b^3-a^3\)\(\Leftrightarrow b\left(20b^2-5ab+4ab-a^2\right)\ge19b^3-a^3\)
\(\Leftrightarrow b\left[5b\left(4b-a\right)+a\left(4b-a\right)\right]\ge19b^3-a^3\)
\(\Leftrightarrow b\left(5b+a\right)\left(4b-a\right)\ge19b^3-a^3\)\(\Leftrightarrow\left(5b^2+ab\right)\left(4b-a\right)\ge19b^3-a^3\)
\(\Leftrightarrow\frac{19b^3-a^3}{ab+5b^2}\le4b-a\)
Tương tự ta có: \(\frac{19c^3-b^3}{cb+5c^2}\le4c-b;\)\(\frac{19a^3-c^3}{ac+5a^2}\le4a-c\)
Cộng từng vế của các BĐT trên, ta được:
\(\text{}\text{}\text{Σ}_{cyc}\frac{19b^3-a^3}{ab+5b^2}\le4\left(a+b+c\right)-\left(a+b+c\right)=3\left(a+b+c\right)\)
Dấu "=" xảy ra khi a = b = c
\(VT-VP=\left(a+c\right)^2\left(a-c\right)^2\left[\frac{1}{\left(a+b\right)\left(ab+5b^2\right)+\left(b+c\right)\left(bc+5c^2\right)}+\frac{1}{\left(a+c\right)\left(ac+5a^2\right)}\right]\)
\(+\frac{\left(a+b\right)\left(b+c\right)\left(ab^2-2abc-5ac^2+5b^3-4b^2c+5bc^2\right)^2}{bc\left(a+5b\right)\left(b+5c\right)\left(a^2b+6ab^2+5b^3+b^2c+6bc^2+5c^3\right)}\ge0\)
