[Toán 9] Bất đẳng thức.

E

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Có:$a^2+b \ge 2a\sqrt{b}$

\Rightarrow $\dfrac{\sqrt{a}}{a^2+b+2b\sqrt{a}} \le \dfrac{sqrt{a}}{2\sqrt{ab}(\sqrt{a}+\sqrt{b}}$

Tương tự \Rightarrow $VT \le \dfrac{\sqrt{a}+\sqrt{b}}{2 \sqrt{ab}. (\sqrt{a}+ \sqrt{b})}= \dfrac{1}{2}. \dfrac{1}{\sqrt{ab}} \le \dfrac{1}{4}(\dfrac{1}{a}+\dfrac{1}{b})=\dfrac{1}{2}$

\Rightarrow đpcm
 
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