Question

Cho x,y \(\ge\) 0 thỏa mãn x^2+y^2=1. Tìm gtnn \(A=\sqrt{5x+4}+\sqrt{5y+4}\)

Answer 1

Do \(x^2+y^2=1\Rightarrow\left\{{}\begin{matrix}\left|x\right|\le1\\\left|y\right|\le1\end{matrix}\right.\) \(\Rightarrow0\le x;y\le1\)

\(\Rightarrow\left\{{}\begin{matrix}x\ge x^2\\y\ge y^2\end{matrix}\right.\) \(\Rightarrow x+y\ge x^2+y^2=1\)

Mặt khác, do \(\left\{{}\begin{matrix}x\ge0\\y\ge0\end{matrix}\right.\) \(\Rightarrow xy\ge0\)

Ta có:

\(A^2=5\left(x+y\right)+8+2\sqrt{25xy+20\left(x+y\right)+16}\)

\(\Rightarrow A^2\ge5.1+8+2\sqrt{25.0+20.1+16}=25\)

\(\Rightarrow A\ge5\) (do \(A>0\))

\(\Rightarrow A_{min}=5\) khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)

Answer 2

Nguyễn Việt Lâm giúp mk nhá, tks bn nhìu ;>>