Cho a, b, c > 0. Chứng minh rằng $\frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2\text{a}}\ge \frac{a^2+b^2+c^2}{3}$

$\frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2\text{a}}=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2\text{ac}}$

Áp dụng bất đẳng thức Cauchy – schwarz ta có:$\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2\text{ac}}\ge \frac{{\left(a^2+b^2+c^2\right)}^2}{a^2+2\text{a}b+b^2+2bc+c^2+2\text{a}c}$

 $\iff \frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2\text{a}}\ge \frac{{\left(a^2+b^2+c^2\right)}^2}{{\left(a+b+c\right)}^2}$                   (1)

Theo hệ quả của bất đẳng thức AM – GM ta có $\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2\text{ac}}\ge \frac{{\left(a^2+b^2+c^2\right)}^2}{a^2+2\text{a}b+b^2+2bc+c^2+2\text{a}c}$

$\iff \frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2\text{a}}\ge \frac{{\left(a^2+b^2+c^2\right)}^2}{{\left(a+b+c\right)}^2}$

a² + b² + c² ≥ ab + bc + ac

⇔ 2(a² + b² + c²) ≥ 2(ab + bc + ac)

⇔ 3(a² + b² + c²) ≥ 2ab + 2bc + 2ac + a² + b² + c²

⇔ 3(a² + b² + c²) ≥ (a + b + c)²                                                              (2)

Từ (1) và (2) suy ra $\frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2\text{a}}\ge \frac{a^2+b^2+c^2}{3}$

Dấu “ = ” xảy ra khi a = b = c > 0

Vậy với a, b, c > 0 thì $\frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2\text{a}}\ge \frac{a^2+b^2+c^2}{3}$ .