Question

cho a,b là 2 số thực dương sao cho a - \(\sqrt{a}\)= \(\sqrt{b}\) -b

tìm giá trị nhỏ nhất của P= \(a^2\)+ \(b^2\) + \(\dfrac{2020}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)

Answer 1

\(a-\sqrt{a}=b-\sqrt{b}\Rightarrow a+b=\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\Rightarrow a+b\le2\)

\(P=a^2+b^2+\dfrac{2020}{\left(\sqrt{a}+\sqrt{b}\right)^2}=a^2+b^2+\dfrac{2020}{\left(a+b\right)^2}\)

\(P\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{2020}{\left(a+b\right)^2}=\dfrac{1}{2}\left(a+b\right)^2+\dfrac{8}{\left(a+b\right)^2}+\dfrac{2012}{\left(a+b\right)^2}\)

\(P\ge2\sqrt{\dfrac{8\left(a+b\right)^2}{2\left(a+b\right)^2}}+\dfrac{2012}{2^2}=507\)

\(P_{min}=507\) khi \(a=b=1\)