Ch $x,y,z$ thực không âm chứng minh $$3xyz+x^3+y^3+z^3 \ge 2 \left[(xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(xz)^{\frac{3}{2}} \right ]$$
Chứng minh $3xyz+x^3+y^3+z^3 \ge 2 \left[(xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(xz)^{\frac{3}{2}} \right ]$
Started By Ispectorgadget, 01-01-2013 - 03:04
popoviciu
#1
Posted 01-01-2013 - 03:04
►|| The aim of life is self-development. To realize one's nature perfectly - that is what each of us is here for. ™ ♫
#2
Posted 01-01-2013 - 06:15
$2\sqrt[2]{(xy)^3} \le xy(x+y)$Ch $x,y,z$ thực không âm chứng minh $$3xyz+x^3+y^3+z^3 \ge 2 \left[(xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(xz)^{\frac{3}{2}} \right ]$$
Suy ra Cần chứng minh :
$$x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
...
- Mai Duc Khai, caybutbixanh, no matter what and 4 others like this
Off vĩnh viễn ! Không ngày trở lại.......
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users