\(a;\left(đk:a,b>0\right)\sqrt{\dfrac{a^2}{b}}+\sqrt{\dfrac{b^2}{a}}=\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\left(đpcm\right)\)
\(b;\) \(bđt\) \(bunhia\Rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2c^2}=c\sqrt{2}\)