Question

Cho x,y không âm thỏa mãn: \(x^2+y^2=1\)

a, CMR: \(1\le x+y\le\sqrt{2}\)

b, Tìm GTLN và GTNN của \(P=\sqrt{1+2x}+\sqrt{1+2y}\)

Answer 1

a.

\(x^2+y^2=1\Rightarrow0\le x;y\le1\)

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

\(x+y\le\sqrt{2\left(x^2+y^2\right)}=\sqrt{2}\)

b.

\(P\le\sqrt{2\left(1+2x+1+2y\right)}\le\sqrt{2\left(2+2\sqrt{2}\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{1+2x}=a\\\sqrt{1+2y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}1\le a;b\le\sqrt{3}\\a^2+b^2=2+2\left(x+y\right)\ge4\end{matrix}\right.\)

\(\left(a-1\right)\left(a-\sqrt{3}\right)\le0\Rightarrow a^2+\sqrt{3}\le a\left(1+\sqrt{3}\right)\Rightarrow a\ge\frac{a^2+\sqrt{3}}{1+\sqrt{3}}\)

Tương tự: \(b\ge\frac{b^2+\sqrt{3}}{1+\sqrt{3}}\)

\(\Rightarrow P=a+b\ge\frac{a^2+b^2+2\sqrt{3}}{1+\sqrt{3}}\ge\frac{4+2\sqrt{3}}{1+\sqrt{3}}=1+\sqrt{3}\)