\(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\)
\(A=\dfrac{a^3}{a^2+3ab+b^2}+\dfrac{b^3}{b^2+3bc+c^2}+\dfrac{c^3}{c^2+3ca+a^2}\)
Áp dụng BĐT Caushy ta có:
\(A\ge\dfrac{a^3}{a^2+3.\dfrac{a^2+b^2}{2}+b^2}+\dfrac{b^3}{b^2+3.\dfrac{b^2+c^2}{2}+c^2}+\dfrac{c^3}{c^2+3.\dfrac{c^2+a^2}{2}+a^2}\)
\(=\dfrac{a^3}{\dfrac{5}{2}\left(a^2+b^2\right)}+\dfrac{b^3}{\dfrac{5}{2}\left(b^2+c^2\right)}+\dfrac{c^3}{\dfrac{5}{2}\left(c^2+a^2\right)}\)
\(=\dfrac{2}{5}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\) (1)
Ta có: \(\dfrac{a^3}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)
Tương tự: \(\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2};\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)
Cộng vế theo vế của các BĐT trên ta có:
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge\dfrac{a+b+c}{2}=\dfrac{3}{2}\left(2\right)\)
Từ (1), (2) suy ra:
\(A\ge\dfrac{2}{5}.\dfrac{3}{2}=\dfrac{3}{5}\left(đpcm\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)