\(y=\sqrt{x^2-2px+2p^2}+\sqrt{x^2-2qx+2q^2}\\ =\sqrt{\left(x-p\right)^2+p^2}+\sqrt{\left(x-q\right)^2+q^2}\left(1\right)\)
Ta thấy
\(\left\{{}\begin{matrix}\sqrt{\left(x-p\right)^2}\ge0\\\sqrt{\left(x-q\right)^2}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x-p\right)^2+p^2}\ge\sqrt{p^2}=\left|p\right|=0\\\sqrt{\left(x-q\right)^2+q^2}\ge\sqrt{q^2}=\left|q\right|=0\end{matrix}\right.\\ \Rightarrow Min\left(1\right)0khi\left\{{}\begin{matrix}x=p=0\\x=q=0\end{matrix}\right.\Leftrightarrow x=p=q=0\)
P/s: Làm hơi tắt nên có mấy chỗ chưa logic lắm '-'
\(y=\sqrt{\left(p-x\right)^2+p^2}+\sqrt{\left(x-q\right)^2+q^2}\ge\sqrt{\left(p-x+x-q\right)^2+\left(p+q\right)^2}\)
\(y\ge\sqrt{\left(p-q\right)^2+\left(p+q\right)^2}=\sqrt{2\left(p^2+q^2\right)}\)
Dấu "=" xảy ra khi \(x=\frac{p+q}{2}\)
