\(\Sigma\dfrac{x^3}{y+z}=\Sigma\dfrac{x^4}{xy+xz}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2.\sqrt{\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2\right)}}=\dfrac{x^2+y^2+z^2}{2}\ge\dfrac{\left(x+y+z\right)^2}{2+2+2}\ge\dfrac{3^2}{6}=\dfrac{3}{2}\Rightarrow min=\dfrac{3}{2}\Leftrightarrow x=y=z=1\)
- Đặt \(A=\dfrac{x^3}{y+z}+\dfrac{y^3}{z+x}+\dfrac{z^3}{x+y}\)
\(=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{yz+yx}+\dfrac{z^4}{zx+zy}\)
- Áp dụng BĐT Caushy-Schwarz, ta có:
\(A=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{yz+yx}+\dfrac{z^4}{zx+zy}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+xz+yz+yx+zx+zy}=\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+zx\right)}\left(1\right)\)- Mặt khác, ta có:
\(\left\{{}\begin{matrix}x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\left(2\right)\\xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\left(3\right)\end{matrix}\right.\)
- Từ (1) , (2), (3) suy ra:
\(A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{\dfrac{\left(x+y+z\right)^4}{9}}{2.\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{\left(x+y+z\right)^2}{6}\ge\dfrac{3^2}{6}=\dfrac{3}{2}\)
- Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
- Vậy \(MinA=\dfrac{3}{2}\)
