Question

cho a,b,c là các số dương. Tính giá trị nhỏ nhất của P= (a+b+c).\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Answer 1

C1: Ap dụng BĐT Caushy-Shwarz dạng Engel: Ta có: a,b,c>0

\(P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\Rightarrow\dfrac{P}{a+b+c}=\dfrac{1^2}{a}+\dfrac{1^2}{b}+\dfrac{1^2}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\)\(\Rightarrow P\ge9\) Dấu bằng xảy ra \(\Leftrightarrow\)\(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Leftrightarrow a=b=c\)

C2: Khai triển

\(P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+1+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+1+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge1+2+1+2+1+2=9\)Dấu "=" xảy ra \(\Leftrightarrow a^2=b^2=c^2\Leftrightarrow a=b=c\)