[Toán 9] Tìm GTNN

  • Thread starter minhducnguyen_2000@yahoo.com.vn
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soicon_boy_9x

$2T= \sum \dfrac{2x^2}{y+z} \geq \sum \dfrac{2x^2}{\sqrt{2(y^2+z^2)}}$

Đặt $\sqrt{y^2+z^2}=a \ \ \ \ \ \sqrt{x^2+z^2}=b \ \ \ \ \ \ \sqrt{x^2+y^2}=c$

$\rightarrow 2T=\sum \dfrac{b^2+c^2-a^2}{\sqrt{2}a}$

$\leftrightarrow 2\sqrt{2}T=\sum \dfrac{b^2}{a} +\sum \dfrac{c^2}{a} - a-b-c$

$\leftrightarrow 2\sqrt{2}T \geq a+b+c=2014$(BCS)

$\leftrightarrow T \geq \dfrac{1007\sqrt{2}}{2}$

Dấu "=" xảy ra $\leftrightarrow x=y=z=...$

Vậy ...
 
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