Bài 1. Cho x, y, z >0 thỏa mãn x+y+z=3. Chứng minh rằng:
$\frac{x^{3}}{y^{3}+8}+\frac{y^{3}}{z^{3}+8}+\frac{z^{3}}{x^{3}+8}\geq \frac{1}{3}\geq \frac{1}{9}+\frac{2}{27}.(xy+yz+zx)$
Sử dụng bất đẳng thức cô-si:
$\frac{x^3}{(y+2)(y^2-2y+4)}+\frac{y+2}{27}+\frac{y^2-2y+4}{27}\geq \frac{x}{3}$
$\frac{y^3}{(z+2)(z^2-2z+4)}+\frac{z+2}{27}+\frac{z^2-2z+4}{27}\geq \frac{y}{3}$
$\frac{z^3}{(x+2)(x^2-2x+4)}+\frac{x+2}{27}+\frac{x^2-2x+4}{27}\geq \frac{z}{3}$
Cộng từng vế các BĐT trên ta có
$\sum \frac{x^3}{y^3+8}+ \frac{x+y+z+6}{27}+\frac{x^2+y^2+z^2-2x-2y-2z+12}{27}\geq \frac{x+y+z}{3}$
$\Rightarrow \sum \frac{x^3}{y^3+8}\geq \frac{4}{9}-\frac{x^2+y^2+z^2}{27}$
$=\frac{4}{9}-\frac{(x+y+z)^2-2(xy+yz+xz)}{27}=\frac{1}{9}+\frac{2}{27}(xy+yz+xz)$