Giải bài 10 trang 107 – SGK môn Đại số lớp 10

Hướng dẫn:

Xét hiệu $\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}-\left( \sqrt{a}+\sqrt{b} \right)$

Ta có:
$\begin{aligned} & \dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}-\left( \sqrt{a}+\sqrt{b} \right)\\&=\dfrac{{{\left( \sqrt{a} \right)}^{3}}+{{\left( \sqrt{b} \right)}^{3}}-\sqrt{ab}\left( \sqrt{a}+\sqrt{b} \right)}{\sqrt{ab}} \\ & =\dfrac{\left( \sqrt{a}+\sqrt{b} \right)\left( a+b-2\sqrt{ab} \right)}{\sqrt{ab}} \\ & =\dfrac{\left( \sqrt{a}+\sqrt{b} \right){{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{\sqrt{ab}}\ge 0\,\,\,\forall a,b>0 \\ \end{aligned}$

Suy ra $\dfrac a {\sqrt{b}}+\dfrac b {\sqrt a}\ge \sqrt a +\sqrt b$