Question

Cần giải gấp

Answer 1

a.

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge\left(\dfrac{1}{\sqrt{a}}.\sqrt{a}+\dfrac{1}{\sqrt{b}}.\sqrt{b}+\dfrac{1}{\sqrt{c}}.\sqrt{c}\right)^2=9\)

\(\Rightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

\(\Rightarrow A\ge3\) 

\(A_{min}=3\) khi \(a=b=c=1\)

b.

\(3B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{4b}+\dfrac{1}{9c}\right)\ge\left(\dfrac{1}{\sqrt{a}}.\sqrt{a}+\dfrac{1}{2\sqrt{b}}.\sqrt{b}+\dfrac{1}{3\sqrt{c}}.\sqrt{c}\right)^2=\dfrac{121}{36}\)

\(\Rightarrow B\ge\dfrac{121}{108}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{18}{11};\dfrac{9}{11};\dfrac{6}{11}\right)\)