Do \(x^2+y^2=1\Rightarrow0\le x;y\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\end{matrix}\right.\) \(\Rightarrow x+y\ge x^2+y^2=1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4+5x}=a\\\sqrt{4+5y}=b\end{matrix}\right.\) \(\left\{{}\begin{matrix}2\le a;b\le3\\a^2+b^2=8+5\left(x+y\right)\ge13\end{matrix}\right.\)
Do \(2\le a;b\le3\Rightarrow\left\{{}\begin{matrix}\left(a-2\right)\left(a-3\right)\le0\\\left(b-2\right)\left(b-3\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2-5a+6\le0\\b^2-5b+6\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a\ge\frac{a^2+6}{5}\\b\ge\frac{b^2+6}{5}\end{matrix}\right.\)
\(\Rightarrow P=a+b\ge\frac{a^2+6}{5}+\frac{b^2+6}{5}=\frac{a^2+b^2+12}{5}\ge\frac{13+12}{5}=5\)
\(\Rightarrow P_{min}=5\) khi \(\left(a;b\right)=\left(2;3\right)\) và hoán vị hay \(\left(x;y\right)=\left(0;1\right)\) và hoán vị
Ủng hộ cách khác
Dễ cm: \(x\ge x^2;y\ge y^2\)
\(P=\sqrt{4+5x}+\sqrt{4+5y}=\sqrt{x+4x+4}+\sqrt{y+4y+4}\ge\sqrt{x^2+4x+4}+\sqrt{y^2+4y+4}=\left|x+2\right|+\left|y+2\right|=x+y+4\ge x^2+y^2+4=1\)"=" khi x;y là hoán vị của (0;1)
