Question

Chứng minh: \(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\), với mọi a > 1

Answer 1

\(\sqrt{a+1}-\sqrt{a-1}=\dfrac{\left(\sqrt{a+1}-\sqrt{a-1}\right)\left(\sqrt{a+1}+\sqrt{a-1}\right)}{\sqrt{a+1}+\sqrt{a-1}}\)

\(=\dfrac{2}{\sqrt{a+1}+\sqrt{a-1}}\)

Mà \(1.\sqrt{a+1}+1.\sqrt{a-1}< \sqrt{\left(1+1\right)\left(a+1+a-1\right)}=2\sqrt{a}\) (dấu "=" của BĐT Bunhia ko xảy ra)

\(\Rightarrow\dfrac{2}{\sqrt{a+1}+\sqrt{a-1}}>\dfrac{2}{2\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

Hay \(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) (đpcm)