Question

Cho a, b, c \(\ge\) 0 thỏa mãn : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\). Tìm GTLN của :

A= \(\dfrac{1}{\sqrt{a^2-ab+b^2}}+\dfrac{1}{\sqrt{b^2-bc+c^2}}+\dfrac{1}{\sqrt{c^2-ca+a^2}}\)

 

Answer 1

Ta có:

\(\sqrt{a^2-ab+b^2}=\sqrt{\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2}\ge\dfrac{1}{2}\left(a+b\right)\)

Tương tự, suy ra BĐT cần chứng minh là:

\(A\le\text{∑}\dfrac{2}{a+b}\)
Áp dụng BĐT Svácxơ, ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Áp dụng 

=> \(A\le\text{∑}\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra ⇔\(a=b=c=1\)