ĐKXĐ: \(0\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{1-x}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}0\le a;b\le1\\a^2+b^2=1\end{matrix}\right.\)
\(A=\sqrt{1+a}+\sqrt{1+b}\le\sqrt{2\left(2+a+b\right)}\le\sqrt{2\left(2+\sqrt{2\left(a^2+b^2\right)}\right)}\)
\(\Rightarrow A\le\sqrt{4+2\sqrt{2}}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\) hay \(x=\frac{1}{2}\)
// Do \(\left\{{}\begin{matrix}0\le a\le1\\0\le b\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a\left(a-1\right)\le0\\b\left(b-1\right)\le0\\ab\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\ge a^2\\b\ge b^2\\ab\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+b\ge a^2+b^2=1\\ab\ge0\end{matrix}\right.\)
\(A^2=2+a+b+2\sqrt{1+a+b+ab}\ge2+1+2\sqrt{1+1+0}=3+2\sqrt{2}\)
\(\Rightarrow A^2\ge\left(\sqrt{2}+1\right)^2\Rightarrow A\ge\sqrt{2}+1\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)
