Question

Với x, y>0, x+y=1. Tìm giá trị nhỏ nhất của biểu thức: \(P=\dfrac{x}{\sqrt{1-x}}+\dfrac{y}{\sqrt{1-y}}\)

Answer 1

\(\sqrt{x\left(1-x\right)}\le\dfrac{1}{2}\left(x+1-x\right)=\dfrac{1}{2}\Rightarrow\sqrt{1-x}\le\dfrac{1}{2\sqrt{x}}\)

\(\Rightarrow\dfrac{1}{\sqrt{1-x}}\ge2\sqrt{x}\Rightarrow\dfrac{x}{\sqrt{1-x}}\ge2x\sqrt{x}\)

\(\Rightarrow P\ge2x\sqrt{x}+2y\sqrt{y}\ge2\sqrt{\left(x^2+y^2\right)\left(\sqrt{x}^2+\sqrt{y}^2\right)}\ge2\sqrt{\dfrac{\left(x+y\right)^2}{2}\left(x+y\right)}=\sqrt{2}\)

\(\Rightarrow P_{min}=\sqrt{2}\) khi \(x=y=\dfrac{1}{2}\)