Question

if a>0 and b>0 prove sqrt of a +sqrt of b >sqrt of a+b

Answer 1

For \(a,b>0\),
\[2\sqrt{ab}>0\]Thus:
\[a+2\sqrt{ab}+b>a+b\]Let \(\sqrt{a}\) and \(v=\sqrt{b}\). Then,\[u^2+2uv+v^2>u^2+v^2\]\[(u+v)^2>u^2+v^2\]\[\sqrt{(u+v)^2}=\sqrt{u^2+v^2}\]\[|u+v|=\sqrt{u^2+v^2}\]Since \(u,v>0\), \(|u+v|=u+v\). Replace \(u\) and \(v\) with
\(a\) and \(b\).

Answer 2

*appropriate expressions for a and b