Áp dụng BĐT Cauchy cho các số không âm, ta có:

\(\dfrac{c}{a}+\dfrac{b}{c}\ge2\sqrt{\dfrac{c}{a}\cdot\dfrac{b}{c}}=2\sqrt{\dfrac{b}{a}}\)

\(\dfrac{b}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{b}{a}\cdot\dfrac{a}{b}}=2\)

Vì \(2\sqrt{\dfrac{b}{a}}\ge2\) nên \(\dfrac{c}{a}+\dfrac{b}{c}\ge\dfrac{b}{a}+\dfrac{a}{b}\) (đpcm)

Câu 1:

\(2\sqrt{\dfrac{3}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)

= \(\sqrt{\dfrac{2^2\cdot3}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)

= \(\sqrt{\dfrac{12}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)

= \(\dfrac{\sqrt{12}\cdot\sqrt{20}}{\left(\sqrt{20}\right)^2}+\dfrac{\sqrt{60}}{\left(\sqrt{60}\right)^2}-\dfrac{\sqrt{15}}{\left(\sqrt{15}\right)^2}\)

= \(\dfrac{\sqrt{240}}{20}+\dfrac{\sqrt{60}}{60}-\dfrac{\sqrt{15}}{15}\)

= \(\dfrac{\sqrt{15}}{5}+\dfrac{\sqrt{15}}{30}-\dfrac{\sqrt{15}}{15}\)

= \(\sqrt{15}\cdot\left(\dfrac{1}{5}+\dfrac{1}{30}-\dfrac{1}{15}\right)\)

= \(\sqrt{15}\cdot\dfrac{1}{6}\) = \(\dfrac{\sqrt{15}}{6}\)

Bài 2:

a)\(\dfrac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}=\dfrac{1}{\sqrt{18}+2\sqrt{2}-2\sqrt{2}}=\dfrac{1}{\sqrt{18}}=\dfrac{\sqrt{18}}{18}=\dfrac{\sqrt{2}}{6}\)

b)\(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{1+2\sqrt{2}+2-3}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=\dfrac{1}{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)\)c) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{3+2\sqrt{6}+2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}{2\left(\sqrt{6}\right)^2}=\dfrac{\sqrt{6}}{12}\cdot\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\)

Trục căn thức ở mẫu:

a) \(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)

b) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}\)