có a>b>0=>\(\sqrt{a}>\sqrt{b}\)
\(\sqrt{a}-\sqrt{b}< \sqrt{a-b}< =>a+b-2\sqrt{ab}< a-b\)
\(< =>2b-2\sqrt{ab}< 0\)
\(< =>-2\sqrt{ab}+2b< 0\)
\(< =>-2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)< 0\)
có \(-2\sqrt{b}< 0\)
mà \(\sqrt{a}>\sqrt{b}=>\sqrt{a}-\sqrt{b}>0\)
=>\(-2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)< 0\)(luôn đúng với \(a>b>0\))
=>\(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
Vì a >b > 0 nên:
\(\sqrt{b}-\sqrt{a}< 0\)
\(< =>2\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)< 0\)
\(< =>2\left(b-\sqrt{ab}\right)< 0\)
\(< =>2b-2\sqrt{ab}< 0\)
\(< =>a-2\sqrt{ab}+b-a+b< 0\)
\(< =>a-2\sqrt{ab}+b< a-b\)
\(< =>\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)
\(=>\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
