Ta có :
\(a+b+c=1\)
\(\Leftrightarrow\) \(\dfrac{a+b+c}{a}=\dfrac{1}{a}\)
\(\Leftrightarrow1+\dfrac{b}{a}+\dfrac{c}{a}=\dfrac{1}{a}\)(1)
Tương tự ta lại có :
\(1+\dfrac{a}{b}+\dfrac{c}{b}=\dfrac{1}{b}\) (2)
\(1+\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{1}{c}\) (3)
Từ 1 ; 2 và 3 :
\(\left\{{}\begin{matrix}1+\dfrac{b}{a}+\dfrac{c}{a}=\dfrac{1}{a}\\1+\dfrac{a}{b}+\dfrac{c}{b}=\dfrac{1}{b}\\1+\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{1}{c}\end{matrix}\right.\)
Cộng vế theo vế ta được :
\(3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
căng thế @Dương Phan Khánh Dương
Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=9\)
Nếu chưa học thì dùng bunyakovsky :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left(\sqrt{\dfrac{1}{a}}^2+\sqrt{\dfrac{1}{b}}^2+\sqrt{\dfrac{1}{c}}^2\right)\ge\left(\dfrac{\sqrt{a}}{\sqrt{a}}+\dfrac{\sqrt{b}}{\sqrt{b}}+\dfrac{\sqrt{c}}{\sqrt{c}}\right)^2=9\)
Em có cách này không biết có đc ko ạ.
Áp dụng BĐT AM-GM cho 3 số: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}=9\)
_ tthnew _
