Question

[Hoang Tran]

cho a,b>0 thỏa mãn a+b≤1.Tìm GTNN của biểu thức

P=\(a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}\)

Answer 1

[Nguyễn Việt Lâm]

\(P\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(\dfrac{4}{a+b}\right)^2\)

\(P\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{8}{\left(a+b\right)^2}=\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2\left(a+b\right)^2}+\dfrac{15}{2\left(a+b\right)^2}\)

\(P\ge\dfrac{1}{2}.2\sqrt{\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2}}+\dfrac{15}{2.1^2}=\dfrac{17}{2}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)