[Toán 9] Tìm GTNN

  • Thread starter minhducnguyen_2000@yahoo.com.vn
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E

eye_smile

$P=\dfrac{1}{a}+\dfrac{1}{b} \ge \dfrac{4}{a+b} \ge \dfrac{4}{2\sqrt{2}}=\sqrt{2}$
 
T

transformers123

Ta có:

$\dfrac{1}{a}+\dfrac{1}{b} \ge 2.\sqrt{\dfrac{1}{ab}}= \dfrac{4}{2\sqrt{ab}} \ge \dfrac{4}{a+b} \ge \dfrac{4}{2\sqrt{2}} =\sqrt{2}$

Dấu "=" xảy ra khi $x=y=\sqrt{2}$
 
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