Question

Giúp mình với ạ:>

Answer 1

Gọi bđt trên là A .

ta có:

\(A=\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

=> \(A\ge3+2+2+2=9\)

Vậy \(A\ge9\) 

Xảy ra bđt khi và chỉ khi a = b = c.