Question

\(Q=\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2.}\)

cho x+y=1 tim min Q

Answer 1

Lâu rồi  hổng thấy ai giải nên giải luôn ak 

Ta có \(5x^2+2xy+2y^2=\left(2x+y\right)^2+\left(x-y\right)^2\ge\left(2x+y\right)^2\Rightarrow\sqrt{5x^2+2xy+2y^2}\ge2x+y.\)

           \(2x^2+2xy+5y^2=\left(x+2y\right)^2+\left(x-y\right)^2\ge\left(x+2y\right)^2\Rightarrow\sqrt{2x^2+2xy+5y^2}\ge x+2y.\)

Suy ra \(Q\ge3\left(x+y\right)=3.1=3\)dấu = xảy ra khi \(\hept{\begin{cases}x+y=1\\x-y=0\end{cases}\Leftrightarrow}x=y=\frac{1}{2}\)