Cho x,y,z > 0 và x + y + z = 1 Tìm Min

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20071006

Tajuu kage bushin no jutsu!

$\sum \frac{x^2(y+z)}{yz}$\geq$ \sum \frac{2x^2\sqrt{yz}}{yz}$
\Leftrightarrow$\sum \frac{x^2(y+z)}{yz}$\geq$ \sum \frac{2x^2}{\sqrt{yz}}$\geq $\sum \frac{4x^2}{y+z}$
\Leftrightarrow$\sum \frac{x^2(y+z)}{yz}$\geq$ \sum \frac{4x^2}{y+z}$\geq$ \frac{(2x+2y+2z)^2}{2(x+y+z)}=2$
\Leftrightarrow$\sum \frac{x^2(y+z)}{yz}$\geq $2$
Dấu "=" xảy ra \Leftrightarrow$x=y=z=\frac{1}{3}$


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