bạn dung bđt a+b >= 2 căn ab ( cô si )  nhé

cách là ghép từng cặp ở vế trái lại

 Ta có: \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\)

\(=\frac{1}{2}\left(\frac{ab}{c}+\frac{bc}{a}\right)+\frac{1}{2}\left(\frac{bc}{a}+\frac{ca}{b}\right)+\frac{1}{2}\left(\frac{ca}{b}+\frac{ab}{c}\right)\)

\(\ge\frac{1}{2}\cdot2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}+\frac{1}{2}\cdot2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}+\frac{1}{2}\cdot2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}\) (Cauchy)

\(=\frac{1}{2}\cdot2b+\frac{1}{2}\cdot2c+\frac{1}{2}\cdot2a\)

\(=a+b+c\)

Dấu "=" xảy ra khi: a = b = c

\(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\)

\(=\frac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a}+\sqrt{b}}=\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left[\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\right]}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2\left(dpcm\right)\)