a) Ta có:
\(+) \sqrt{25 + 9}=\sqrt{34}\).
\(+) \sqrt{25} + \sqrt{9}=\sqrt{5^2}+\sqrt{3^2}=5+3\)
\(=8=\sqrt{8^2}=\sqrt{64}\).
Vì \(34<64\)
Vậy \(\sqrt{25 + 9}<\sqrt{25} + \sqrt{9}\)
b) Với \(a>0,b>0\), ta có
\(+)\, (\sqrt{a + b})^{2} = a + b\).
\(+) \,(\sqrt{a} + \sqrt{b})^{2}= (\sqrt{a})^2+ 2\sqrt a .\sqrt b +(\sqrt{b})^2\)
\( = a +2\sqrt{ab} + b\)
\(=(a+b) +2\sqrt{ab}\).
Vì \(a > 0,\ b > 0\) nên \(\sqrt{ab} > 0 \Leftrightarrow 2\sqrt{ab} >0\)
\(\Leftrightarrow (a+b) +2\sqrt{ab} > a+b\)
\(\Leftrightarrow (\sqrt{a}+\sqrt{ b})^2 > (\sqrt{a+b})^2\)
\(\Leftrightarrow \sqrt{a}+\sqrt{b}>\sqrt{a+b}\) (đpcm)
