Question

[Gay\]

Cho 3 số dương a,b,c. Tìm GTNN của:
\(P=\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(P\ge\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\right)\)

\(P\ge\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\right)\)

\(P\ge\left(a+b+c\right)^2\left(\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{\dfrac{1}{3}\left(a+b+c\right)^2}\right)=30\)

\(P_{min}=30\) khi \(a=b=c\)