Question

CMR : \(\dfrac{1}{\sqrt{1\cdot199}}+\dfrac{1}{\sqrt{2\cdot198}}+\dfrac{1}{\sqrt{3\cdot197}}+...+\dfrac{1}{\sqrt{199\cdot1}}>1,99\)

Answer 1

Theo BĐT Cô-si ta có: với hai số dương a, b: \(\sqrt{ab}\le\dfrac{a+b}{2}\Rightarrow\dfrac{1}{\sqrt{a.b}}\ge\dfrac{2}{a+b}\)

Dấu "=" xảy ra khi a=b

Áp dụng vào bài toán:

\(\dfrac{1}{\sqrt{1.199}}+\dfrac{1}{\sqrt{2.198}}+...+\dfrac{1}{\sqrt{199.1}}>\dfrac{2}{1+199}+\dfrac{2}{2+198}+...+\dfrac{2}{199+1}\)

\(VT>\dfrac{2}{200}+\dfrac{2}{200}+...+\dfrac{2}{200}\) (199 thừa số)

\(VT>\dfrac{2.199}{200}=1.99\) (đpcm)