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vf321Best ResponseYou've already chosen the best response.0
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 \(b\).
 one year ago

vf321Best ResponseYou've already chosen the best response.0
*appropriate expressions for a and b
 one year ago
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