Question

Cho a+b+c=1. Tìm GTNN

A= \(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+\left(c+\dfrac{1}{c}\right)^2\)

Answer 1

Áp dụng BĐT cauchy schwar dưới dạng engel ta có :

\(\dfrac{\left(a+\dfrac{1}{a}\right)^2}{1}+\dfrac{\left(b+\dfrac{1}{b}\right)^2}{1}+\dfrac{\left(c+\dfrac{1}{c}\right)^2}{1}\ge\dfrac{\left(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{3}\)

Tiếp tục áp dụng BĐT trên :

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

\(\Rightarrow\)\(\dfrac{\left(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{3}=\dfrac{\left(1+9\right)^2}{3}=\dfrac{100}{3}\)

Vậy GTNN của \(A=\dfrac{100}{3}\)

Dấu \("="\) xảy ra khi \(a=b=c=\dfrac{1}{3}\)