Giải thích các bước giải:
Ta có:
$\dfrac{a^3}{a^2+b^2}=\dfrac{a(a^2+b^2)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}$
Chứng minh tương tự ta có:
$\begin{cases}\dfrac{b^3}{b^2+c^2} \ge b-\dfrac{c}{2}\\ \dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\end{cases}$
$\rightarrow \dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2} +\dfrac{c^3}{c^2+a^2}\ge a-\dfrac{b}{2}+b-\dfrac{c}{2}+c-\dfrac{a}{2}$
$\rightarrow \dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2} +\dfrac{c^3}{c^2+a^2}\ge\dfrac{1}{2}(a+b+c)=\dfrac{3}{2}$