[toán 9] chúng minh

E

eye_smile

Ta có: $x^4+yz \ge 2x^2\sqrt{yz}$

\Rightarrow $\dfrac{x^2}{x^4+yz} \le \dfrac{x^2}{2x^2\sqrt{yz}}=\dfrac{1}{2\sqrt{yz}}$

Tương tự với 2 số còn lại \Rightarrow $BT \le \dfrac{1}{2\sqrt{yz}}+\dfrac{1}{2\sqrt{xy}}+\dfrac{1}{2\sqrt{xz}} \le \dfrac{1}{2}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})=\dfrac{1}{2}(\dfrac{xy+yz+zx}{xyz}) \le \dfrac{1}{2}.\dfrac{x^2+y^2+z^2}{xyz} \le \dfrac{3}{2}$
 
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