Question

Cho a,b,c > 0thoar mãn a+b+c=1. Tìm GTNN của biểu thức:\(P=\dfrac{9}{2\left(ab+bc+ca\right)}+\dfrac{2}{a^2+b^2+c^2}\)

Answer 1

\(P=2\left(\dfrac{4}{2ab+2ac+2bc}+\dfrac{1}{a^2+b^2+c^2}\right)+\dfrac{1}{2}\left(\dfrac{1}{ab+ac+bc}\right)\)

\(\Rightarrow P\ge\dfrac{2.\left(2+1\right)^2}{2ab+2ac+2bc+a^2+b^2+c^2}+\dfrac{1}{2}.\dfrac{1}{\dfrac{\left(a+b+c\right)^2}{3}}\)

\(\Rightarrow P\ge\dfrac{18}{\left(a+b+c\right)^2}+\dfrac{3}{2}.\dfrac{1}{\left(a+b+c\right)^2}=18+\dfrac{3}{2}=\dfrac{39}{2}\)

\(\Rightarrow P_{min}=\dfrac{39}{2}\) khi \(a=b=c=\dfrac{1}{3}\)