Question

Chứng minh bất đẳng thức sau:

\(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) với a>1

Answer 1

\(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) <=> \(\left(\dfrac{1}{\sqrt{a}}\right)^2< \left(\sqrt{a+1}-\sqrt{a-1}\right)^2\)

<=> \(\dfrac{1}{a}< \left(a+1\right)+\left(a-1\right)-2\sqrt{a^2-1}\)

<=> \(2\sqrt{a^2-1}< 2a-\dfrac{1}{a}\)

<=> \(4\left(a^2-1\right)< 2\left(2a-\dfrac{1}{a}\right)^2\) <=> \(\dfrac{1}{a^2}>0\)

Vậy \(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) với mọi a ≥ 0=> đpcm.