[toán 8] đại số

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T

transformers123

Đặt $P=\dfrac{4}{x+y+z}-\dfrac{1}{3(xy+yz+zx)}$

Ta có: $(x+y+z)^2 \ge 3(xy+yz+zx)$

$\iff \dfrac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2] \ge 0$ (luôn đúng)

Dấu "=" xảy ra khi $x=y=z$

Ta có: $P=\dfrac{4}{x+y+z}-\dfrac{1}{3(xy+yz+zx)}$

$\iff P \le \dfrac{4}{x+y+z}-\dfrac{1}{(x+y+z)^2}$

$\iff P \le \dfrac{4}{a}-\dfrac{1}{a^2}$ (với $a=x+y+z$)

$\iff P \le \dfrac{4a-1}{a^2}$

$\iff P \le \dfrac{-4a^2+4a-1}{a^2}+4$

$\iff P \le \dfrac{-(2a-1)^2}{a^2}+4$

$\iff P \le 4$

Dấu "=" xảy ra khi $a= \dfrac{1}{2} \Longrightarrow x=y=z=\dfrac{1}{6}$
 
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