Question

\(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\)

Tìm GTNN của:

S=\(\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)

Answer 1

áp dụng bất đẳng thức mincopski ta có :

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

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}=\sqrt{3^2+\left(\dfrac{9}{3}\right)^2}=3\sqrt{2}\)

\(\Rightarrow GTNN\) của \(S\) là \(3\sqrt{2}\) dấu "=" xảy ra khi \(a=b=c=1\)