Bài làm:
Ta có: \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}\)
\(=\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\)
Áp dụng Bất đẳng thức Cauchy (AM-GM), ta được:
\(\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\)\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{a}{b}.\frac{b}{a}}\)
\(=2.\sqrt{1}+2.\sqrt{1}+2.\sqrt{1}\)\(=2+2+2\)\(=6\)
=> \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge6\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{a}{b}=\frac{b}{a}\\\frac{c}{a}=\frac{a}{c}\\\frac{b}{c}=\frac{c}{b}\end{cases}\Rightarrow a=b=c=1}\)
Vậy \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=6\)khi \(a=b=c=1\)
Học tốt!!!!
Theo giả thiết : \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=6\)
\(< =>\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}=6\)
\(< =>\frac{ac+bc}{c^2}+\frac{ba+ca}{a^2}+\frac{cb+ba}{b^2}=6\)
Ta có : \(VT=\frac{ac+bc}{c^2}+\frac{ba+ca}{a^2}+\frac{cb+ba}{b^2}\)
\(=\frac{ac}{c^2}+\frac{bc}{c^2}+\frac{ba}{a^2}+\frac{ca}{a^2}+\frac{cb}{b^2}+\frac{ba}{b^2}\)
\(\ge6\sqrt[6]{\frac{a^2c^2b^2c^2b^2a^2}{a^4b^4c^4}}=6\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c\)
Xin chém cách khác ạ =))
