CMR Với x,y,z > 0 thì:
$\frac{{xyz\left( {x + y + z + \sqrt {x^2 + y^2 + z^2 } } \right)}}{{\left( {x^2 + y^2 + z^2 } \right)\left( {xy + yz + zx} \right)}} \le \frac{{3 + \sqrt 3 }}{9}$
$\frac{{xyz\left( {\sum x + \sqrt {\sum {x^2 } } } \right)}}{{\sum {x^2 \sum {xy} } }}$
Bắt đầu bởi NLT, 12-04-2012 - 20:31
#1
Đã gửi 12-04-2012 - 20:31
- Mai Duc Khai và WhjteShadow thích
GEOMETRY IS WONDERFUL !!!
Some people who are good at calculus think that they will become leading mathematicians. It's funny and stupid.
Nguyễn Lâm Thịnh
#2
Đã gửi 12-04-2012 - 22:37
Ta có
$\frac{xyz(x+y+z+\sqrt{x^2+y^2+z^2})}{(x^2+y^2+z^2)(xy+yz+zx)}=\frac{xyz(x+y+z)}{(x^2+y^2+z^2)(xy+yz+zx)}+\frac{xyz}{\sqrt{x^2+y^2+z^2}(xy+yz+zx)}\leq \frac{xyz(x+y+z)}{(xy+yz+zx)^2}+\frac{xyz}{\sqrt3.\sqrt[3]{xyz}.3.\sqrt[3]{x^2y^2z^2}}\leq \frac{1}{3}+\frac{1}{3\sqrt3}$
=>đpcm
$\frac{xyz(x+y+z+\sqrt{x^2+y^2+z^2})}{(x^2+y^2+z^2)(xy+yz+zx)}=\frac{xyz(x+y+z)}{(x^2+y^2+z^2)(xy+yz+zx)}+\frac{xyz}{\sqrt{x^2+y^2+z^2}(xy+yz+zx)}\leq \frac{xyz(x+y+z)}{(xy+yz+zx)^2}+\frac{xyz}{\sqrt3.\sqrt[3]{xyz}.3.\sqrt[3]{x^2y^2z^2}}\leq \frac{1}{3}+\frac{1}{3\sqrt3}$
=>đpcm
- nthoangcute, NLT và hamdvk thích
“There is no way home, home is the way.” - Thich Nhat Hanh
#3
Đã gửi 13-04-2012 - 17:18
CMR Với x,y,z > 0 thì:
$\frac{{xyz\left( {x + y + z + \sqrt {x^2 + y^2 + z^2 } } \right)}}{{\left( {x^2 + y^2 + z^2 } \right)\left( {xy + yz + zx} \right)}} \le \frac{{3 + \sqrt 3 }}{9}$
$\frac{{xyz\left( {x + y + z + \sqrt {x^2 + y^2 + z^2 } } \right)}}{{\left( {x^2 + y^2 + z^2 } \right)\left( {xy + yz + zx} \right)}}\leq \frac{{(x+y+z)\left( {x + y + z + \sqrt {x^2 + y^2 + z^2 } } \right)}}{9{\left( {x^2 + y^2 + z^2 } \right)}} \leq \frac{(x+y+z)^2+\frac{(x+y+z)^2}{\sqrt{3}}}{3(x+y+z)^2}= \frac{3+\sqrt{3}}{9}$
- nthoangcute và NLT thích
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