Áp dụng BĐT Cauchy cho các cặp số dương, ta có: \(VT=\Sigma\frac{a}{\sqrt{b^3+1}}=\Sigma\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\)
\(\ge\Sigma\frac{a}{\frac{\left(b+1\right)+\left(b^2-b+1\right)}{2}}=\Sigma\frac{2a}{b^2+2}=\Sigma\left(a-\frac{ab^2}{b^2+2}\right)\)
\(=\Sigma\left(a-\frac{2ab^2}{b^2+b^2+4}\right)\ge\Sigma\left(a-\frac{2ab^2}{3\sqrt[3]{4b^4}}\right)\)\(=\Sigma\left[a-\frac{a\sqrt[3]{2b^2}}{3}\right]=\Sigma\left[a-\frac{a\sqrt[3]{2.b.b}}{3}\right]\)
\(\ge\Sigma\left[a-\frac{a\left(2+b+b\right)}{9}\right]\)\(=\left(a+b+c\right)-\frac{2\left(a+b+c\right)}{9}-\frac{2\left(ab+bc+ca\right)}{9}\)
\(=\frac{7\left(a+b+c\right)}{9}-\frac{2\left(ab+bc+ca\right)}{9}\)\(\ge\frac{7\left(a+b+c\right)}{9}-\frac{2.\frac{\left(a+b+c\right)^2}{3}}{9}=2\)
Đẳng thức xảy ra khi a = b = c = 2
