[Đại 8] Chứng minh

T

transformers123

Câu 2:
Ap1 dụng bđt Cauchy, ta có:
$\begin{cases}\dfrac{bc}{a}+\dfrac{ca}{b} \ge 2\sqrt{\dfrac{abc^2}{ab}} = 2c\\\dfrac{ca}{b}+\dfrac{ab}{c} \ge 2\sqrt{\dfrac{a^2bc}{bc}} = 2a\\\dfrac{ab}{c}+\dfrac{bc}{a} \ge 2\sqrt{\dfrac{ab^2c}{ac}} = 2b\end{cases} \rightarrow 2(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}) \ge 2(a+b+c)$
$\Longrightarrow \mathfrak{dpcm}$
Dấu "=" xảy ra khi $a=b=c$
 
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T

transformers123

Câu 1:
$x^4+y^4+z^2 \ge 2x(xy^2-x+z+1)$
$\iff x^4+y^4+z^2-2x^2y^2+2x^2-2xz-2x \ge 0$
$\iff x^4-2x^2y^2+y^4+x^2-2xz+x^2+x^2-2x+1 \ge 0$
$\iff (x^2-y^2)^2+(x-z)^2+(x-1)^2 \ge 0$ (luôn đúng)
Dấu "=" xảy ra khi $\begin{cases}x^2-y^2=0\\x-z=0\\x-1=0\end{cases} \iff \begin{cases}x=z=1\\y=\pm 1\end{cases}$
 
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