Chủ đề này có 1 trả lời

[CaoHoangAnh]

Cho $x,y,z> 0 ;xyz=1$

CM $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+z)(1+x)}+\frac{z^3}{(1+y)(1+x)}$$\geq \frac{3}{4}$

[lequangnghia]

Cho $x,y,z> 0 ;xyz=1$

CM $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+z)(1+x)}+\frac{z^3}{(1+y)(1+x)}$$\geq \frac{3}{4}$

Ta có $\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8}\geq \frac{3}{4}x$

 $\frac{y^{3}}{(1+z)(1+x)}+\frac{1+z}{8}+\frac{1+x}{8}\geq \frac{3}{4}y$

 $\frac{z^{3}}{(1+y)(1+x)}+\frac{1+y}{8}+\frac{1+x}{8}\geq \frac{3}{4}z$

Cộng 3 bđt theo vế ta thu được $\sum \frac{x^{3}}{(1+y)(1+z)}\geq \frac{1}{2}(x+y+z)-\frac{3}{4}\geq \frac{1}{2}3\sqrt[3]{xyz}-\frac{3}{4}\geq \frac{3}{4}$